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This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- According to Bohr model electrons having different energies belong to different levels having different values of n. So, their angular momenta will be different, as

L=nh/2 $\pi L\propto$ n

This is a Short Answer Type Questions as classified in NCERT Exemplar

Metallic character- Tendency to lose electrons.

Non-Metallic character- Tendency to accept electrons.

Across the period the metallic character decreases whereas non-metallic character increases across the period.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- If proton had a charge (+4/3)e and electron a charge (-3/4)e, then the Bohr formula for the H-atom remain same, since the Bohr formula involves only the product of the charges which remain constant for given values of charges.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- The transition of an electron from a higher energy to a lower energy level can appears in the form of electromagnetic radiation because electrons interact only electromagnetically.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- On removing one electron from He4 and He3, the energy levels, as worked out on the basis of Bohr model will be very close as both the nuclei are very heavy as compared to electron mass.Also after removing one electron from He4 and He3 atoms contain one electron and are hydrogen like atoms.

This is a Short Answer Type Questions as classified in NCERT Exemplar

(a) As the effective nuclear charge increases and shielding effect decreases across the periods thus the electronegativity increases on moving from left to right in the periodic table.

(b) As the number of shells increases down the group thus its ionisation enthalpy decreases down the group.

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This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- the reason behind this is the binding energy or we can say the mass defect.

BE= mass defect (931MeV ). So whatever energy is liberated in fission and fusion changes the mass slighthly. That why there is a defect in mass

The deviation in the ionization enthalpy of some elements from the general trend can be explained by the points as given below:-

(i) The fully filled and half filled orbital provide extra stability due to the symmetry

(ii) The effective nuclear charge

(iii) The e^- - e^- repulsion which lead to instability

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- when r0=1A^o

Let $\epsilon =2+\delta$

F= $\frac{{kq}_1{q}_2}{r^{2+\delta }}{R}_0^{\delta }$

Where $\frac{{kq}_1{q}_2}{r^2}$ = ( ${1.6\times {10}^{-19})}^2\times 9\times {10}^9=23.04\times {10}^{-29}N/m2$

Let p= $23.04\times {10}^{-29}N/m2$

Electrostatic force is balanced by centripetal force

Mv²/r or v²= $\frac{p{R}_0^{\delta }}{m{r}^{1+\delta }}$

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This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- m_p= 10^-6

M_pc²=10^{-6 $\times electronmass\times c^2$}

 = 0.8 $\times {10}^{-19}J$

Wavelength associated with both of them is same

$\frac{h}{m_{p}c}=\frac{hc}{m_p{c}^2}=\frac{{10}^{-34}\times 3\times {10}^8}{0.8\times {10}^{-19}}=4\times {10}^{-7}m$

U(r)=- $\frac{e^2}{4\pi {\epsilon }_o}\frac{\mathrm{e}\mathrm{x}\mathrm{p}?(-\lambda t)}{r}$

Mvr=h

V=h/mr

Mv²/r=( $\frac{e^2}{4\pi {\epsilon }_o}$ ) ( $\frac{1}{r^2}+\frac{\lambda }{r}$ )

$\frac{h^2}{m{r}^3}=\left(\frac{e^2}{4\pi {\epsilon }_o}\right)\left(\frac{1}{r^2}-\frac{\lambda }{r}\right)$

$\frac{h^2}{m}=\frac{e^2}{4\pi {\epsilon }_o}{r}_B$

$kinecticwillbedoubleofgroundstateso$

13.6 $\times 2=27.2eV$

 This is a Short Answer Type Questions as classified in NCERT Exemplar

(i) S

The ionization enthalpy increases across the period and decreases down the group. N possesses a half filled 2p orbital which provides it extra stability due to symmetry.

(ii)P

The non-metallic character increases across the period

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- the energy of nth state E_n=-Z²R $\frac{1}{n^2}$  where R is constant and Z=24

The energy release ina transition from 2 to 1 $?E=Z^{2}R\left(1-\frac{1}{4}\right)=\frac{3}{4}{Z}^{2}R$

The energy required to eject a n=4 electron is E₄= Z²R1/16

So kinetic energy of auger electron is, KE= Z²R (3/4-1/16)=1/16Z²R

= $\frac{11}{16}\times 24\times 24\times 13.6eV=5385.6eV$

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This is a Short Answer Type Questions as classified in NCERT Exemplar

Exothermic reaction- The reaction in which heat is released

For example- CaO + CO₂ → CaCO₃ + Heat

Endothermic reaction- The reaction in which heat is absorbed.

For example- 2NH₃ + heat → N₂ +3H₂

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- in this type of situation centripetal force is balanced by electrostatic force

So mv²/r= -ke²/r²

According to bohr postuates

Mvr = h when n=1

On solving n  $\frac{h^2}{m^2{r}^2}\frac{1}{r}={ke}^2/r^2$

So after solving r= 0.51A⁰

So potential energy  $\frac{{ke}^2}{r}$ = -27.2eV , KE= mv²/2

= $\frac{1}{2}m\frac{h^2}{r^2}=\frac{h}{2m{r}^2}=+13.6eV$

But if R

If R>>r the electron moves inside the sphere with radius r’

Charge inside r’⁴=e $\frac{{r\text{'}}^3}{R^3}$

So r’ = $\frac{h^2}{m}\frac{k}{e^2}\frac{R^3}{{r\text{'}}^3}$

So r’⁴= 0.51A⁰R³

= 510(A⁰)⁴