Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • When the wavelength of radiation falling on metal is changed from  to 

When the wavelength of radiation falling on metal is changed from 500\; nm to 200\; nm the maximum kinetic energy of the photoelectrons becomes there times larger. The wave function (in eV)  of the metal is close to
 
Option: 1 0.61
Option: 2 1.02
Option: 3 0.52
Option: 4 0.81

Answers (1)

best_answer

\mathrm{K.E.}=\mathrm{hv}-\Phi

\begin{array}{l} \frac{\mathrm{hc}}{\lambda_{1}}=\phi+(KE)_{1} \cdot(\mathrm{i}) \\ \\ \frac{\mathrm{hc}}{\lambda_{2}}=\phi+(KE)_{2}(\mathrm{ii}) \end{array}

from (i) - (ii)

\mathrm{hc}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)= (KE_2-KE_1)=3KE_1-KE_1=2KE_1\\ KE_1=\frac{\mathrm{hc}}{2}\left(\frac{1}{\lambda_{2}}-\frac{1}{\lambda_{1}}\right)

Now, \phi =\frac{hc}{\lambda _1}-KE_1

So, \phi =\frac{hc}{\lambda _1}-(\frac{\mathrm{hc}}{2}\left(\frac{1}{\lambda_{2}}-\frac{1}{\lambda_{1}}\right))= \frac{3hc}{2\lambda _1}-\frac{hc}{2\lambda _2}\\\Rightarrow \phi =\frac{3*1240}{2*500}-\frac{1240}{2*200}=3.7-3.1=0.6 \ eV

Posted by

avinash.dongre

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Advanced 2024: Chemical engineering cut-offs at IITs; know opening, closing ranks
anam-khan

IIT JEE Mains Result 2024 Session 2: How is overall merit list prepared?
anam-khan

JEE Mains Session 2 result 2024 soon; List of top NITs in India

Latest Question