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

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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

$W=2.36 eV$

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