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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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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

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

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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
$hf_{1}-W=eV_{1}----(3)$
$hf_{2}-W=eV_{2}----(4)$
Where $V_{1}=5 V_{2}\;$ (given)-----(5)
Using 3,4,5
$\large\frac{h f_{1}-W}{e}=5 (\large\frac{h f_{2}-W}{e})$
$4W=h\;(5 f_{2}-f_{1})$
$W=\large\frac{h}{4}\;[5f_{2}-f_{1}]-----(6)$
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
 

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