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Q.

The stopping potential V₀ (in volt) as a function of frequency (v) for sodium emitter is shown in figure. The work function of sodium from the data plotted in figure would be : $(h = 6 .63 \times 10^{- 34} J - s, charge = 1 .6 \times 10^{- 19} C)$

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a

1.02 eV

b

1.66 eV

c

1.95 eV

d

2.12 eV

answer is B.

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Detailed Solution

$hv = {hv}_{0} + \frac{1}{2} {mu}^{2} = {hv}_{0} + {eV}_{0} = w . f . + {eV}_{0}$
$∴$ Stopping potential is eV₀
$∴ If V_{0} = 0, then v_{0} = 4 \times 10^{14} Hz$
$∴$ Work function = hv₀
$= \frac{6.63 \times 10^{- 34} \times 4 \times 10^{14}}{1.6 \times 10^{- 19}} eV = 1.66 eV$

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