Q60 / 120 · NEET 2011 (Mains) · Physics

The threshold frequency for a photo-sensitive metal is $3.3 \times 10^{14} \mathrm{~Hz}$. If light of frequency $8.2 \times 10^{14} \mathrm{~Hz}$ is incident on this metal, the cut-off voltage for the photo-electric emission is nearly

  • Correct answer
    A
    $2 \mathrm{~V}$
  • B
    $3 \mathrm{~V}$
  • C
    $5 \mathrm{~V}$
  • D
    $1 \mathrm{~V}$

Solution

$$ \text { } \begin{aligned} V_0 & =\frac{E-V}{e} \\ & =\frac{h\left(v-v_0\right)}{e} \\ & =\frac{6.62 \times 10^{-34}\left(8.2 \times 10^{14}-3.3 \times 10^{14}\right)}{1.6 \times 10^{-19}} \\ & =\frac{6.62 \times 10^{-34}}{1.6} \times 4.9 \times 10^{33} \\ & =\frac{6.62 \times 4.9 \times 10^{-1}}{1.6} \end{aligned} $$ $$ V_0=2 \text { volt } $$

Hence, the correct answer is option: 'A'.

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