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Q.

A radiation of wavelength λ illuminates a metal and ejects photoelectrons of maximum kinetic energy of 1 eV. Another radition of wavelength λ/3, ejects photoelectrons of maximum kinetic energy of 4eV. What will be the work function of metal ?

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a

1 eV

b

2 eV

c

0.5 eV

d

3 eV

answer is C.

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Detailed Solution

hv = w + KE

$KE=\frac{hc}{\lambda }-w$

$K{{E}_{1}}=\frac{hc}{{{\lambda }_{1}}}-w$

given $K{{E}_{1}}=1eV$

${{\lambda }_{1}}=\lambda$

$1=\frac{hc}{\lambda }-w$

$\Rightarrow \frac{hc}{\lambda }=1+w$ ... (1)

$\Rightarrow K{{E}_{2}}=\frac{hc}{{{\lambda }_{2}}}-w$

given $K{{E}_{2}}=4eV$

${{\lambda }_{2}}=\frac{\lambda }{3}$

$4=\frac{hc}{\left( \frac{\lambda }{3} \right)}-w$

$4=\frac{3hc}{\lambda }-w$

$\frac{3hc}{\lambda }=4+w$ ... (2)

$\frac{eq\,\,(2)}{eq\,\,(1)}=(3)$

$3=\frac{4+w}{1+w}$

3+3w = 4+w

2w = 1

w = 1.5

work function = 0.5 eV

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