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Stopping Potential Revisited

A mercury arc...

Question

A mercury arc lamp provides $0.1 W$ of ultra-violet radiation at a wavelength of $λ = 2537 \circ A$ only. The photo tube (cathode of photo electric device) consists of potassium and has an effective area of $4 c m^{2}$ . The cathode is located at a distance of $1 m$ from the radiation source. The work function for potassium is $ϕ_{0} = 2.2 e V$ .
What is the cut off potential $V_{0}$ ?

A

26.9 V

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B

2.69 V

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C

1.35 V

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D

5.33 V

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Solution

The correct option is B $2.69 V$
Cut-off potential $V_{0}$ found by photoelectric equation i.e.
$E_{p} - ϕ = e V_{0}$
We have $E_{p}$ as
$E_{p} = \frac{h c}{λ} = \frac{12.43 \times 10^{- 7}}{2537 \times 10^{- 10}} \approx 4.9 eV$
$ϕ = 2.2 eV$
Putting these values in photoelectric equation we get
$2.7 eV = e V_{0}$
$\Rightarrow V_{0} = 2.7 V$

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