Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
0 votes

In an experiment of photoelectric effect , singly ionized helium is excited electronically to different energy levels. The emission spectra of helium produced is incident on a photo - electric plate in photocell .When helium is excited to fourth energy - level than the observed stopping potential of photocell is found to be five times stopping potential when hydrogen is excited to third level . Then the work function of material of photo-electric plate is

$(a)\;4.12 eV\qquad(b)\;2.36eV\qquad(c)\;2.6V\qquad(d)\;1.2 V$

Can you answer this question?

1 Answer

0 votes
Answer : (b) $\;2.36 eV$
Explanation :
Energy of photon of maximum frequency of emission spectra is given by
for helium : $\;f_{1}=(2)^2 R c [\large\frac{1}{1^2}-\large\frac{1}{4^2}]----(1)$
for hydrogen : $\;f_{2}= R c [\large\frac{1}{1^2}-\large\frac{1}{3^2}]----(2)$
using einstein's equation of photo electric we obtain
Where $V_{1}=5 V_{2}\;$ (given)-----(5)
Using 3,4,5
$\large\frac{h f_{1}-W}{e}=5 (\large\frac{h f_{2}-W}{e})$
$4W=h\;(5 f_{2}-f_{1})$
Putting values of $\;f_{1}\;$ & $\;f_{2}\;$ from (1) & (2) in (6) ,
$W=\large\frac{h}{4}\;[(1-\large\frac{1}{9})5 R c -4 (1-\large\frac{1}{16}) R c]$
$=\large\frac{h}{4} R c \;[\large\frac{8 \times5}{9}-\large\frac{15}{4}]=0.694 R h c$
Putting Rhc =13.6eV we get
$W=2.36 eV$
answered Feb 25, 2014 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App