Q157 / 160 · TS EAMCET 2007 · Physics

The work function of the nickel is $5 \mathrm{eV}$. When a light of wavelength $2000 Å$ falls on it, it emits photoelectrons in the circuit. Then the potential difference necessary to stop the fastest electrons emitted is (given $\left.h=6.67 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$

  • A
    $1.0 \mathrm{~V}$
  • B
    $1.75 \mathrm{~V}$
  • Correct answer
    C
    $1.25 \mathrm{~V}$
  • D
    $0.75 \mathrm{~V}$

Solution

Stopping potential is given by $\begin{aligned} e V_0 & =\frac{h c}{\lambda}-\phi \\ \therefore \quad e V_0 & =\frac{6.67 \times 10^{-34} \times 3 \times 10^8}{2 \times 10^{-7}}\end{aligned}$ $-5 \times 1.6 \times 10^{-19}$ $\begin{aligned} & =10.005 \times 10^{-19}-8 \times 10^{-19} \\ & =2.005 \times 10^{-19}\end{aligned}$ or $\quad \begin{aligned} V_0 & =\frac{2.005 \times 10^{-19}}{1.6 \times 10^{-19}} \text { volt } \\ & =1.25 \mathrm{~V}\end{aligned}$

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

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