Light of wavelength $330 n m$ falling on a piece of metal ejects electrons with sufficient energy with required voltage $V_{0}$ to prevent them from reaching a collector. In the same set up, light of wavelength $220 n m$ ejects electrons which require twice the voltage $V_{0}$ to stop them in reaching a collector. The numerical value of voltage $V_{0}$ is

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$\frac{8}{15} V$

$\frac{15}{8} V$

$\frac{15}{16} V$

$\frac{16}{15} V$

answer is C.

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Detailed Solution

$e V_{0} = \frac{h c}{λ_{1}} - W . . . (i)$
Similarly, $e (2 V_{0}) = \frac{h c}{λ_{2}} - W . . . (i i)$
Subtracting $E q .$ $(i)$ from $E q . (i i)$ , we get
$e V_{0} = \frac{h c (λ_{1} - λ_{2})}{λ_{1} λ_{2}}$
$∴ V_{0} = \frac{h c (λ_{1} - λ_{2})}{e λ_{1} λ_{2}} = \frac{(6.63 \times 10^{- 34}) (3 \times 10^{8}) (110)}{1.6 \times 10^{- 19} \times 330 \times 220 \times 10^{- 9}} \approx 1.8 v o l t$