Ask & Answer: India's Largest Education Community

Chemistry NCERT Exemplar Solutions Class 12th Chapter Four Chemical Kinetics

Guide-Level 15

Yes, candidates have to fill out the application form for Faculty of Allied Health Sciences MRIIRS online. Aspirants can apply by paying an application fee of INR 1,200. Below are the steps to apply for admission:

Step 1: Visit the official website of MRIIRS.

Step 2: Click on 'Apply Now' on the Home page.

Step 3: Then Click on 'Apply Now' in front of the preferred MRIIRS institute/ faculty.

Step 4: Register with the basic details to generate login credentials.

Step 5: Login to fill out the application form and upload the documents.

Step 6: Pay the application fee to submit the form.

0 0

New Question

10 months ago

Consider a first order gas phase decomposition reaction given below:

A(g) → B(g) + C(g)

The initial pressure of the system before decomposition of A waspt . After lapse of time ‘t’, total pressure of the system increased by x units and became ‘t’ the rate constant k for the reaction is given as _________.

0 Follower 11 Views

alok kumar singh

Contributor-Level 10

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option B

Given:

A (g) → B (g) + C (g)

p_i= Initial pressure

(Time) t = 0

A (g) → B (g) + C (g)

p_i→0atm + 0atm

t, (p_i - x)atm

p_t = (p_i - x)atm + x + x = p_i + x

p_A = (p_i- x)

The value of x changes when it is substituted.

p_A = p_i- (p_t - p_i) = 2p_i- p_t

K = $\frac{2.303}{t}$ log $\frac{p_{i}}{2 p_{i} - p_{t}}$

0 0

New Question

10 months ago

13.28 Consider the D–T reaction (deuterium–tritium fusion)

${}_{1}^{2}{H + {}_{1}^{3}{H \to {}_{2}^{4}{H e + n}}}$

(a) Calculate the energy released in MeV in this reaction from the data:

m( ${}_{1}^{2}{H)}$ =2.014102 u

m( ${}_{1}^{3}{H)}$ =3.016049 u

(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction?

(Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzmann’s constant, T = absolute temperature.)

0 Follower 8 Views

Faculty of Allied Health Sciences, Manav Rachna International Institute of Research and Studies Colleges in Haryana

Contributor-Level 10

13.28 The equation for deuterium-tritium fusion is given as:

${}_{1}^{2}{H + {}_{1}^{3}{H \to {}_{2}^{4}{H e + n}}}$

It is given that

Mass of ( ${}_{1}^{2}{H), m_{1}}$ = 2.014102 u

Mass of ( ${}_{1}^{3}{H)}$ , $m_{2} =$ 3.016049 u

Mass of ( ${}_{2}^{4}{H e), m_{3}}$ = 4.002603 u

Mass of ( ${}_{0}^{1}{n), m_{4}}$ = 1.008665 u

Q-value of the given D-T reaction is:

Q = $[m_{1} + m_{2} - m_{3} - m_{4}] c^{2}$

= $[2.014102 + 3.016049 - 4.002603 - 1.008665] c^{2} u$

= 0.018883 $c^{2} u$

= 17.59 MeV

Radius of the deuterium and tritium, r $\approx 2.0 f m$ = 2 $\times 10^{- 15}$ m

Distance between the centers of the nucleus when they touch each other,

d = r +r = 4 $\times 10^{- 15}$ m

Charge on the deuterium and tritium nucleus = e

Hence the repulsive potential energy between the two nuclei is given as:

V = $\frac{e^{2}}{4 π ?_{0} d}$

Where,

$?_{0}$ = permittivity of free space

It is given that $\frac{1}{4 π ?_{0}}$ = 9 $\times 10^{9}$ N $m^{2} C^{- 2}$

Hence, V = $\frac{{(1.6 \times 10^{- 19})}^{2}}{4 \times 10^{- 15}}$ $\times$ 9 $\times 10^{9}$ = 5.76 $\times 10^{- 14}$ J =

...more

13.28 The equation for deuterium-tritium fusion is given as:

${}_{1}^{2}{H + {}_{1}^{3}{H \to {}_{2}^{4}{H e + n}}}$

It is given that

Mass of ( ${}_{1}^{2}{H), m_{1}}$ = 2.014102 u

Mass of ( ${}_{1}^{3}{H)}$ , $m_{2} =$ 3.016049 u

Mass of ( ${}_{2}^{4}{H e), m_{3}}$ = 4.002603 u

Mass of ( ${}_{0}^{1}{n), m_{4}}$ = 1.008665 u

Q-value of the given D-T reaction is:

Q = $[m_{1} + m_{2} - m_{3} - m_{4}] c^{2}$

= $[2.014102 + 3.016049 - 4.002603 - 1.008665] c^{2} u$

= 0.018883 $c^{2} u$

= 17.59 MeV

Radius of the deuterium and tritium, r $\approx 2.0 f m$ = 2 $\times 10^{- 15}$ m

Distance between the centers of the nucleus when they touch each other,

d = r +r = 4 $\times 10^{- 15}$ m

Charge on the deuterium and tritium nucleus = e

Hence the repulsive potential energy between the two nuclei is given as:

V = $\frac{e^{2}}{4 π ?_{0} d}$

Where,

$?_{0}$ = permittivity of free space

It is given that $\frac{1}{4 π ?_{0}}$ = 9 $\times 10^{9}$ N $m^{2} C^{- 2}$

Hence, V = $\frac{{(1.6 \times 10^{- 19})}^{2}}{4 \times 10^{- 15}}$ $\times$ 9 $\times 10^{9}$ = 5.76 $\times 10^{- 14}$ J = $\frac{5.76 \times 10^{- 14}}{1.6 \times 10^{- 19}}$ eV = 360 keV

Hence, 360 keV of kinetic energy is needed to overcome the Coulomb repulsion between the two nuclei.

It is given that

KE = 2 $\times \frac{3}{2} k T$

Where

k = Boltzmann constant = 1.38 $\times 10^{- 23}$ $m^{2} k g s^{- 2} K^{- 1}$

T = Temperature required to trigger the reaction

Therefore T = $\frac{K E}{3 k}$ = $\frac{5.76 \times 10^{- 14}}{3 \times 1.38 \times 10^{- 23}}$ = 1.39 $\times 10^{9}$ K

Hence the gas must be heated to 1.39 $\times 10^{9}$ K to initiate the reaction.

less

0 0

New Question

10 months ago

What is the application fee at Faculty of Allied Health Sciences MRIIRS?

0 Follower 3 Views

CHSL SSC CHSL Selection Process

Guide-Level 15

Aspirants applying for admission at Faculty of Allied Health Sciences MRIIRS need to pay an application fee of INR 1,200. Candidates can pay the application fee via any of the online payment modes. Aspirants should keep the fee receipt with them for any future reference.

0 0

New Question

10 months ago

What is SSC CHSL selection process?

0 Follower 31 Views

Upasana Hazarika

Contributor-Level 6

The SSC CHSL selection process comprises two tiers- Tier 1 and Tier 2. Candidates need to clear both the tiers to get shortlisted for appointment. Have a look at the details below:

Tier I: Online objective-type exam
Tier II: Online objective-type exam and skill test

The final allocation of candidates is made by indenting recruiting bodies.

0 0

New Question

10 months ago

NCERT Class 12 Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Four

Consider Fig. 4.1 and mark the correct option.

A. Activation energy of forward reaction is E₁+ E₂and product is less stable than reactant.

B. Activation energy of forward reaction is E₁+ E₂and product is more stable than reactant.

C. Activation energy of both forward and backward reaction isE₁ + E₂and reactant is more stable than product.

D. Activation energy of backward reaction is E₁and product is more stable than reactant.

0 Follower 10 Views

alok kumar singh

Contributor-Level 10

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option A

Activation Energy → The amount of energy required to overcome the obstacle and generate a product

The activation energy of a forward reaction can be seen in the [E_af = E₁ + E₂] (This is an endothermic reaction.

Therefore, [E_af > E_ab]

The energy of the product is high, while the energy of the reactant is low.

The lower the energy, the more stable and positive the situation becomes.

Δ? = posistive

0 0

New Question

10 months ago

13.27 Consider the fission of ${}_{92}^{238}U$ by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are ${}_{58}^{140}{C e}$ and ${}_{44}^{99}{R u})$ . Calculate Q for this fission process. The relevant atomic and particle masses are

m( ${}_{92}^{238}U)$ =238.05079 u

m( ${}_{58}^{140}{C e}$ ) =139.90543 u

m( ${}_{44}^{99}{R u})$ = 98.90594 u

0 Follower 4 Views

How expensive is Tokyo for students?

Contributor-Level 10

13.27 In the fission of ${}_{92}^{238}U$ , $β^{-}$ 10 particles decay from the parent nucleus. The nuclear reaction can be written as:

${}_{92}^{238}U + {}_{0}^{1}{n \to {}_{58}^{140}{C e +}} {}_{44}^{99}{R u + 10 {}_{- 1}^{0}e}$

It is given that:

Mass of a ${}_{92}^{238}U n u c l e u s, m_{1}$ = 238.05079 u

Mass of a ${}_{58}^{140}{C e} n u c l e u s, m_{2}$ =139.90543 u

Mass of a ${}_{44}^{99}{R u} n u c l e u s, m_{3}$ = 98.90594 u

Mass of a neutron ${}_{0}^{1}n$ , $m_{4}$ = 1.00865 u

Q value of the above equation,

Q = $[m^{'} ({}_{92}^{238}U) + m' ({}_{0}^{1}{n) - m^{'} (} {}_{58}^{140}{C e) - m^{'} ({}_{44}^{99}{R u) - 10 m_{e}}}] c^{2}$

Where

m’ = represents the corresponding atomic masses of the nuclei.

$m^{'} ({}_{92}^{238}U)$ = $m_{1} - 92 m_{e}$

m’( ${}_{58}^{140}{C e)}$ = $m_{2} - 58 m_{e}$

m’( ${}_{44}^{99}{R u)}$ = $m_{3} - 44 m_{e}$

m’( ${}_{0}^{1}{n)} = m_{4}$

Substituting these values, we get

Q = $[m_{1} - 92 m_{e} + m_{4} - m_{2} + 58 m_{e} - m_{3} + 44 m_{e} - 10 m_{e}] c^{2}$

= $[m_{1} + m_{4} - m_{3} - m_{2}] c^{2}$

= $[238.05079 + 1.00865 - 98.90594 - 139.90543] c^{2}$ u

=0.24807 $c^{2}$ u

= 231.077 MeV

0 0

New Question

10 months ago

Japan Study Abroad

0 Follower 2 Views

Nikita Shekhar

Beginner-Level 3

Tokyo is arguably the most expensive region to study in Japan, with average monthly cost of living (including food & other utilities) ranging between Yen 57,000-Yen 90,000 (INR 33.9 K-INR 53.6 K). Food, Transportation & other everyday utilities account for about INR 5 K, while the rest comes under rent & personal expenses. The table below gives a detailed breakdown of the living expenses students may incur over a month in Tokyo:

Product	Cost in Yen	Cost in INR
Rice (5 kg)	2,650 yen	INR 1.5 K
White Bread (1 kg)	511 yen	INR 304
Milk (1,000 ml)	253 yen	INR 150
Eggs (10 eggs)	264 yen	INR 157
Apples (1 kg)	1,039 yen	INR 618
Cabbage (1 kg)	178 yen	INR 106
Carbonated drink (1 liter)	236 yen	INR 140
Hamburger	233 yen	INR 138
Gasoline (1 liter)	176 yen	INR 104
Toilet paper (1,000 m)	799 yen	INR 475
Hair cut	3,737 yen	INR 2.2 K
Taxi (4 km)	1,693 yen	INR 1 K
Total Cost	9,110 Yen	INR 5.4 K

Source: Portal Site of Official Statistics of Japan (Retail Price Survey of August 2024)

0 0

New Question

10 months ago

Faculty of Allied Health Sciences, Manav Rachna International Institute of Research and Studies Colleges in Haryana

What should be my MRNAT score to get scholarship at Faculty of Allied Health Sciences MRIIRS?

0 Follower 3 Views

National Institute of Pharmaceutical Education and Research, Ahmedabad

Guide-Level 15

MRIIRS provides scholarships to students based on their MRNAT score. Students with a very high score can also get 100% fee waiver. Below are the details for the MRNAT scholarship available for Faculty of Allied Health Sciences MRIIRS students:

MRNAT Score	'Utkarsh'	'Uttam'
90% and above	100% tuition fee waiver	100% tuition fee waiver
80% and above but below 90%	50% tuition fee waiver	25% tuition fee waiver
70% and above but below 80%	25% tuition fee waiver	10% tuition fee waiver

0 0

New Question

10 months ago

Can I get direct admission at National Institute of Pharmaceutical Education and Research, Ahmedabad?

0 Follower 3 Views

saurya snehal

Contributor-Level 10

Generally, the National Institute of Pharmaceutical Education and Research does not accept direct admissions. Candidates must fulfill the requirements set forth by the university in order to be accepted into any of the programs offered. The procedure can be completed online. Candidates must pass a multi-step admissions procedure that consists of application, entrance, selection, and admission confirmation.
There is no provision of direct admission to NIPER, Ahmedabad.

0 0

New Question

10 months ago

13.26 Under certain circumstances, a nucleus can decay by emitting a particle more massive than an $α -$ particle. Consider the following decay processes:

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

${}_{88}^{223}{R a \to {}_{86}^{219}{R n + {}_{2}^{4}{H e}}}$

Calculate the Q-values for these decays and determine that both are energetically allowed.

0 Follower 4 Views

How many vacancies are announced in SSC CHSL 2025?

Contributor-Level 10

13.26 For the emission of ${}_{6}^{14}C$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{82}^{209}{P b}$ , $m_{2}$ = 208.98107 u

Mass of ${}_{6}^{14}C$ , $m_{3}$ = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ - $m_{3}$ ) $c^{2}$

= (223.01850 - 208.98107 - 14.00324) $c^{2}$ u

= 0.03419 $c^{2}$ u

= 0.03419 MeV = 31.848 MeV

Hence, the Q-value of the nuclear reaction is 31.848 MeV, since the value is positive, the reaction is energetically allowed.

For the emission of ${}_{2}^{4}{H e}$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{86}^{219}{R n + {}_{2}^{4}{H e}}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{86}^{219}{R n}$ , $m_{2}$ = 219.00948 u

Mass of ${}_{2}^{4}{H e}$ , $m_{3}$ = 4.00260 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ &n

...more

13.26 For the emission of ${}_{6}^{14}C$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{82}^{209}{P b}$ , $m_{2}$ = 208.98107 u

Mass of ${}_{6}^{14}C$ , $m_{3}$ = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ - $m_{3}$ ) $c^{2}$

= (223.01850 - 208.98107 - 14.00324) $c^{2}$ u

= 0.03419 $c^{2}$ u

= 0.03419 MeV = 31.848 MeV

Hence, the Q-value of the nuclear reaction is 31.848 MeV, since the value is positive, the reaction is energetically allowed.

For the emission of ${}_{2}^{4}{H e}$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{86}^{219}{R n + {}_{2}^{4}{H e}}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{86}^{219}{R n}$ , $m_{2}$ = 219.00948 u

Mass of ${}_{2}^{4}{H e}$ , $m_{3}$ = 4.00260 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ - $m_{3}$ ) $c^{2}$

= (223.01850 - 219.00948 - 4.00260) $c^{2}$ u

= 6.42 $\times 10^{- 3} c^{2}$ u

= 6.42 $\times 10^{- 3} \times 931.5$ MeV = 5.980 MeV

Hence, the Q-value of the nuclear reaction is 5.98 MeV, since the value is positive, the reaction is energetically allowed.

less

0 0

New Question

10 months ago

SSC CHSL Vacancies SSC CHSL

0 Follower 26 Views

Rupali Pruthi

Contributor-Level 10

The Staff Selection Commission (SSC) has announced a total of 3131 vacancies in the SSC CHSL 2025. These vacancies will be filled for the posts of Lower Division Clerk (LDC) and Data Entry Operator (DEO) in central government ministries and departments. Candidates who want to get recruited through the CHSL (10+2) exam need to apply online. The SSC CHSL applications were invited online.

0 0

New Question

10 months ago

Institute of Forensic Science, Mumbai Forensic Science

What is the admission criteria in Institute of Forensic Science, Mumbai for B.Sc in Forensic science?

0 Follower 4 Views

New Question

10 months ago

Faculty of Allied Health Sciences, Manav Rachna International Institute of Research and Studies Colleges in Haryana

Can I get scholarship at Faculty of Allied Health Sciences MRIIRS based on Class 12 score?

0 Follower 3 Views

Guide-Level 15

Yes, candidates can get scholarship at Faculty of Allied Health Sciences MRIIRS based on their Class 12 percentage. MRIIRS provides Merit-based scholarships to students pursuing a course from the university. Below is the detailed criteria:

Marks/ CGPA in Qualifying Examination	Tuition Fee Waiver
98% and above	100%
95% to 97.99%	50%
90% to 94.99%	25%
85% to 89.99%	10%

0 0

New Question

10 months ago

13.24 The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei ${}_{20}^{41}{C a}$ and from the following data:

m( ${}_{20}^{40}{C a)}$ = 39.962591 u

m( ${}_{20}^{41}{C a)}$ = 40.962278 u

m( ${}_{13}^{26}{A l)}$ = 25.986895 u

m( ${}_{13}^{27}{A l)}$ = 26.981541 u

0 Follower 8 Views

Contributor-Level 10

13.24 If a neutron ${}_{0}^{1}n)$ is removed from ${}_{20}^{41}{C a}$ , the corresponding reaction can be written as:

${}_{20}^{41}{C a} \to$ ${}_{20}^{40}{C a}$ + ${}_{0}^{1}n$

The separation energies are

For ${}_{20}^{41}{C a}$ : Separation energy = 8.363007 MeV

For ${}_{13}^{27}{A l}$ : Separation energy = 13.059 MeV

It is given that

m( ${}_{20}^{40}{C a)}$ = 39.962591 u

m( ${}_{20}^{41}{C a)}$ = 40.962278 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m ${}_{20}^{40}{C a)} +$ ( m( ${}_{0}^{1}n) -$ m( ${}_{20}^{41}{C a)}$

= 39.962591 + 1.008665 - 40.962278

= 8.978 $\times 10^{3}$ u

=8.363007 MeV

For ${}_{13}^{27}{A l}$ , the neutron removal reaction can be written as

${}_{13}^{27}{A l} \to$ ${}_{13}^{26}{A l}$ + ${}_{0}^{1}n$

It is given that

m( ${}_{13}^{26}{A l)}$ = 25.986895 u

m( ${}_{13}^{27}{A l)}$ = 26.981541 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m( ${}_{13}^{26}{A l)} +$ m(

...more

13.24 If a neutron ${}_{0}^{1}n)$ is removed from ${}_{20}^{41}{C a}$ , the corresponding reaction can be written as:

${}_{20}^{41}{C a} \to$ ${}_{20}^{40}{C a}$ + ${}_{0}^{1}n$

The separation energies are

For ${}_{20}^{41}{C a}$ : Separation energy = 8.363007 MeV

For ${}_{13}^{27}{A l}$ : Separation energy = 13.059 MeV

It is given that

m( ${}_{20}^{40}{C a)}$ = 39.962591 u

m( ${}_{20}^{41}{C a)}$ = 40.962278 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m ${}_{20}^{40}{C a)} +$ ( m( ${}_{0}^{1}n) -$ m( ${}_{20}^{41}{C a)}$

= 39.962591 + 1.008665 - 40.962278

= 8.978 $\times 10^{3}$ u

=8.363007 MeV

For ${}_{13}^{27}{A l}$ , the neutron removal reaction can be written as

${}_{13}^{27}{A l} \to$ ${}_{13}^{26}{A l}$ + ${}_{0}^{1}n$

It is given that

m( ${}_{13}^{26}{A l)}$ = 25.986895 u

m( ${}_{13}^{27}{A l)}$ = 26.981541 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m( ${}_{13}^{26}{A l)} +$ m( ${}_{0}^{1}n) -$ m( ${}_{13}^{27}{A l)}$

= 25.986895 + 1.008665 - 26.981541

= 0.014019 u

=13.0586985 MeV

less

0 0

New Question

10 months ago

13.15 The Q value of a nuclear reaction A + b $\to$ C + d is defined by

Q = [ $m_{A} + m_{b}$ – $m_{C}$ – $m_{d}$ ] $c^{2}$

where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.

(i) ${}_{1}^{1}H$ + ${}_{1}^{3}H$ $\to$ ${}_{1}^{2}H$ + ${}_{1}^{2}H$

(ii) ${}_{6}^{12}C$ + ${}_{6}^{12}C$ $\to$ ${}_{10}^{20}{N e}$ + ${}_{2}^{4}{H e}$

Atomic masses are given to be

m ( ${}_{1}^{2}H$ ) = 2.014102 u

m ( ${}_{1}^{3}H$ ) = 3.016049 u

m ( ${}_{6}^{12}C$ ) = 12.000000 u

m ( ${}_{10}^{20}{N e}$ ) = 19.992439 u

0 Follower 12 Views

Faculty of Allied Health Sciences, Manav Rachna International Institute of Research and Studies Colleges in Haryana

Contributor-Level 10

13.15 The given nuclear reaction is

${}_{1}^{1}H$ + ${}_{1}^{3}H$ $\to$ ${}_{1}^{2}H$ + ${}_{1}^{2}H$

Atomic mass

m ( ${}_{1}^{1}H$ ) = 1.007825 u

m ( ${}_{1}^{2}H$ ) = 2.014102 u

m ( ${}_{1}^{3}H$ ) = 3.016049 u

The Q-value of the reaction can be written as:

Q = $[m ({}_{1}^{1}H) + m ({}_{1}^{3}H) - 2 m ({}_{1}^{2}H)] c^{2}$

= $[1.007825 + 3.016049 - 2 \times 2.014102] c^{2}$

= (-4.33 $\times 10^{- 3}$ ) $c^{2}$

But 1 u = 931.5 MeV/ $c^{2}$

Q = -4.0334 MeV

The negative Q-value of the reaction shows that the reaction is endothermic.

The given nuclear reaction is

${}_{6}^{12}C$ + ${}_{6}^{12}C$ $\to$ ${}_{10}^{20}{N e}$ + ${}_{2}^{4}{H e}$

Atomic mass

m ( ${}_{6}^{12}C$ ) = 12.000000 u

m ( ${}_{10}^{20}{N e}$ ) = 19.992439 u

m ( ${}_{2}^{4}{H e}$ ) = 4.002603 u

The Q-value of this reaction is given as:

Q = $[2 m ({}_{6}^{12}C) - m ({}_{10}^{20}{N e}) - m ({}_{2}^{4}{H e})] c^{2}$

= $[2 \times 12.0 - 19.992439 - 4.002603] c^{2}$

=4.958 $\times 10^{- 3} c^{2}$ u

=4.958 $\times 10^{- 3} \times 931.5$

=4.6183 MeV

...more

13.15 The given nuclear reaction is

${}_{1}^{1}H$ + ${}_{1}^{3}H$ $\to$ ${}_{1}^{2}H$ + ${}_{1}^{2}H$

Atomic mass

m ( ${}_{1}^{1}H$ ) = 1.007825 u

m ( ${}_{1}^{2}H$ ) = 2.014102 u

m ( ${}_{1}^{3}H$ ) = 3.016049 u

The Q-value of the reaction can be written as:

Q = $[m ({}_{1}^{1}H) + m ({}_{1}^{3}H) - 2 m ({}_{1}^{2}H)] c^{2}$

= $[1.007825 + 3.016049 - 2 \times 2.014102] c^{2}$

= (-4.33 $\times 10^{- 3}$ ) $c^{2}$

But 1 u = 931.5 MeV/ $c^{2}$

Q = -4.0334 MeV

The negative Q-value of the reaction shows that the reaction is endothermic.

The given nuclear reaction is

${}_{6}^{12}C$ + ${}_{6}^{12}C$ $\to$ ${}_{10}^{20}{N e}$ + ${}_{2}^{4}{H e}$

Atomic mass

m ( ${}_{6}^{12}C$ ) = 12.000000 u

m ( ${}_{10}^{20}{N e}$ ) = 19.992439 u

m ( ${}_{2}^{4}{H e}$ ) = 4.002603 u

The Q-value of this reaction is given as:

Q = $[2 m ({}_{6}^{12}C) - m ({}_{10}^{20}{N e}) - m ({}_{2}^{4}{H e})] c^{2}$

= $[2 \times 12.0 - 19.992439 - 4.002603] c^{2}$

=4.958 $\times 10^{- 3} c^{2}$ u

=4.958 $\times 10^{- 3} \times 931.5$

=4.6183 MeV

The positive sign shows that the reaction is exothermic.

less

0 0

New Question

10 months ago

What is the cost of studying BSc at Faculty of Allied Health Sciences MRIIRS?

0 Follower 4 Views

Guide-Level 15

The total tuition fee for BSc/ BSc (Hons) at Faculty of Allied Health Sciences MRIIRS is INR 5.2 lakh. This information is sourced from official website/ sanctioning body and is subject to change. Aspirants who are shortlisted for admission have to pay the first installment of the fee to confirm their seat. The mode of payment is as specified by the university. Besides, aspirants also need to pay a registration fee of INR 1,200 in the start to apply for BSc/ BSc (Hons) programme.

0 0

New Question

10 months ago

13.13 The radionuclide ${}_{6}^{11}C$ decays according to ${}_{6}^{11}{C \to {}_{5}^{11}{B + e^{+} + v :}} T_{1 / 2}$ =20.3 min. The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: m ( ${}_{6}^{11}{C)}$ = 11.011434 u and m ( ${}_{5}^{11}B$ ) = 11.009305 u, calculate Q and compare it with the maximum energy of the positron emitted.

0 Follower 4 Views

Faculty of Allied Health Sciences, Manav Rachna International Institute of Research and Studies Colleges in Haryana

Contributor-Level 10

13.13 The given values are

m ( ${}_{6}^{11}{C)}$ = 11.011434 u and m ( ${}_{5}^{11}B$ ) = 11.009305 u

The given nuclear reaction:

${}_{6}^{11}{C \to {}_{5}^{11}{B + e^{+} + v}}$

Half life of ${}_{6}^{11}C$ nuclei, $T_{1 / 2}$ =20.3 min

The maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (ΔQ) of the nuclear masses of the ${}_{6}^{11}C$

ΔQ = $[m' ({}_{6}^{11}{C)} - {m' ({}_{5}^{11}B) + m_{e}}] c^{2} \dots \dots \dots \dots \dots (1)$

where

$m_{e}$ = Mass of an electron or positron = 0.000548 u

c = speed of the light

m’ = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 $m_{e}$ in the case of ${}_{6}^{11}C$ and 5 $m_{e}$ in the case of ${}_{5}^{11}B$ .

Hence the equation (1) reduces to

ΔQ = $[m ({}_{6}^{11}{C)} - m ({}_{5}^{11}B) - {2 m}_{e}] c^{2}$

=&nb

...more

13.13 The given values are

m ( ${}_{6}^{11}{C)}$ = 11.011434 u and m ( ${}_{5}^{11}B$ ) = 11.009305 u

The given nuclear reaction:

${}_{6}^{11}{C \to {}_{5}^{11}{B + e^{+} + v}}$

Half life of ${}_{6}^{11}C$ nuclei, $T_{1 / 2}$ =20.3 min

The maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (ΔQ) of the nuclear masses of the ${}_{6}^{11}C$

ΔQ = $[m' ({}_{6}^{11}{C)} - {m' ({}_{5}^{11}B) + m_{e}}] c^{2} \dots \dots \dots \dots \dots (1)$

where

$m_{e}$ = Mass of an electron or positron = 0.000548 u

c = speed of the light

m’ = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 $m_{e}$ in the case of ${}_{6}^{11}C$ and 5 $m_{e}$ in the case of ${}_{5}^{11}B$ .

Hence the equation (1) reduces to

ΔQ = $[m ({}_{6}^{11}{C)} - m ({}_{5}^{11}B) - {2 m}_{e}] c^{2}$

= $[11.011434 - 11.009305 - 2 \times 0.000548] c^{2}$ u

=1.033 $\times 10^{- 3} c^{2}$ u

But 1 u = 931.5 MeV/ $c^{2}$

ΔQ = 0.962 MeV

The value of $Δ Q$ is almost comparable to the maximum energy of the emitted positron.

less

0 0

New Question

10 months ago

What is the eligibility for BSc at Faculty of Allied Health Sciences MRIIRS?

0 Follower 3 Views