Search Colleges, Courses, Exams, Questions and Articles

Ask a query on 8879100846 Ask Login | Sign Up

 

 

› Ask & Answer › Question

11.30 Light of intensity 10–5 W m–2 falls on a sodium photo-cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?

0 3 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

  • P

    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

    11.30 Intensity of the incident light, I = 10 - 5  W m - 2

    Surface area of the sodium photocell, A = 2 c m 2  = 2 × 10 - 4 m 2

    The incident power of the light, P = I × A  = 10 - 5 × 10 - 4 m 2 W = 2 × 10 - 9  W

    Work function of the metal, ∅ 0 = 2 eV = 2 × 1.6 × 10 - 19  J = 3.2 × 10 - 19  J

    Number of layers of sodium that absorbs the incident energy, n = 5

    The effective atomic area of a sodium atom, A e  = 10 - 20 m 2

    Hence, the number of conduction electrons in n layers is given by:

    n’ = n × A A e  = 5 × 2 × 10 - 4 10 - 20  = 1 × 10 17

    Since the incident power is uniformly absorbed by all the electrons continuously, the amount of energy absorbed per second per electron is

    E = P n '  = 2 × 10 - 9 1 × 10 17  = 2 × 10 - 26  J/

    ...more

Similar Questions for you

Given below are two statements:

Statement I : Davission-Germer experiment establishes, the wave nature of electrons.

Statement II : If electrons have wave nature, they can interfere and show diffraction.

In the light of the above statements choose the correct answer from the option given below:
R
Raj Pandey

Based on theory

Radiation from hydrogen gas excited to first excited state is used for illuminating certain metallic plate. When the same plate is exposed to the radiation from some unknown hydrogen like gas excited to the same level it is found that de-Broglie wavelength of fastest photoelectron has decreased by 2.3 times. It is given that energy corresponding to longest wavelength of Lyman series of the unknown gas is 3 times the ionization energy of hydrogen gas (13.6eV). The work function (in eV) of the metallic plate is ______. [Take (2.3)2 = 5.25]
V
Vishal Baghel

z² × (13.6) (1 - ¼) = 3 × (13.6)
z = 2 . (i)
h/√2mk? = (1/2.3) × h/√2mk?
=> k? = (2.3)²k? = 5.25k? (ii)
Now, k? = E? - Φ
k? = E? - Φ = z²E? - Φ
∴ k? /k? = (10.2 - Φ)/ (4 × 10.2 - Φ) = 1/5.25
=> Φ = 3eV

A metal exposed to light of wavelength 800nm and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500nm is used. The work function of the metal is: (Take hc = 1230 eV –nm).
V
Vishal Baghel

  K . E 1 = 1 2 3 0 8 0 0 − ? - (i)

  K . E 2 = 2 K . E 1 = 1 2 3 0 5 0 0 − ? - (ii)

from (i) & (ii)

? = 0 . 6 1 5  ev

When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is υ 1 . When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes υ 2 . If υ 2 = x υ 1 , the value of x will be __________.             
V
Vishal Baghel

hu = hu0 + K.E

Cases u = 2u0

h2u0 = hu0 + K.E1

K.E1 = hu0

1 2 m v 1 2 = h υ 0

v 1 2 = 2 h υ 0 m  

v 1 = 2 h υ 0 m - (1)      

Now, cases 2

h 5u0 = hu0 + k.E2

k.E2 = 4hu0

1 2 m v 2 2 = 4 h υ 0

v2 =     8 h υ 0 m   - (2)

v 2 v 1 = 8 2 = 2          

v2 = 2v1

Kindly consider the following

f(x)={2x2−3x+2x−2,x≠25,x=2
A
alok kumar singh

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

Sol:

f(x)=2x2−3x−2x−2                                      =2x2−4x+x−2x−2=2x(x−2)+1(x−2)x−2                                       =(2x+1)(x−2)x−2=2x+1                           limx→2−f(x)=2x+1=limh→02(2−h)+1=4+1=5                          limx→2+f(x)=2x+1=limh→02(2+h)+1=4+1=5                           limx→2f(x)=5As   limx→2−f(x)=limx→2+f(x)=limx→2f(x)=5Hence,  f(x)  is  continuous  at  x=2.

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 687k Reviews
  • 1800k Answers

Learn more about...

Share Your College Life Experience

Write a Review

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Ask Now
Community GuidelinesUser Points System
QuestionsDiscussionsTags
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.

Need guidance on career and education? Ask our experts

Characters 0/140

The Answer must contain atleast 20 characters.

Add more details

Characters 0/300

The Answer must contain atleast 20 characters.

Keep it short & simple. Type complete word. Avoid abusive language. Next

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post