In the stydy of a photoelectric effect the graph between the stopping potential V and frequency v of the incident radiation on two different metals P and Q is shown below:
(i) Which one of the two metals has higher threshold frequency?
(ii) Determine the work function of the metal which has greater value.
(iii) Find the maximum kinetic energy of electron emitted by light of frequency $8 \times 1 0^{14} H z$ for this metal.

Question

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Answer 1

(i) Threshold frequency of P is $3 \times 1 0^{14} H z .$
Threshold frequency of Q is $6 \times 1 0^{14} H z .$
Clearly Q has higher threshold frequency.
(ii) Work function of metal Q, $ψ_{0} = h v_{0}$
$= (6.6 \times 1 0^{- 34}) \times 6 \times 1 0^{14} J$
$= \frac{3 9 . 6 \times 1 0 ^{- 20}}{1 . 6 \times 1 0 ^{- 19}} e V = 2.5 e V$
(iii) Maximum kinetic energy, $K_{m a x} = h v - h v_{0}$
$= h (v - v_{0})$
$= 6.6 \times 1 0^{- 34} (8 \times 1 0^{14} - 6 \times 1 0^{14})$
$= 6.6 \times 1 0^{- 34} \times 2 \times 1 0^{14} J$
$= \frac{6 . 6 \times 1 0 ^{- 34} \times 2 \times 1 0 ^{14} J}{1 . 6 \times 1 0 ^{- 19}} e V$
$∴ K_{m a x} = 0.83 e V$

Answer 2

(i) Threshold frequency of P is

3 \times 1 0^{14} H z .

Threshold frequency of Q is

6 \times 1 0^{14} H z .

Clearly Q has higher threshold frequency.
(ii) Work function of metal Q,

ψ_{0} = h v_{0}

= (6.6 \times 1 0^{- 34}) \times 6 \times 1 0^{14} J

= \frac{3 9 . 6 \times 1 0 ^{- 20}}{1 . 6 \times 1 0 ^{- 19}} e V = 2.5 e V

(iii) Maximum kinetic energy,

K_{m a x} = h v - h v_{0}

= h (v - v_{0})

= 6.6 \times 1 0^{- 34} (8 \times 1 0^{14} - 6 \times 1 0^{14})

= 6.6 \times 1 0^{- 34} \times 2 \times 1 0^{14} J

= \frac{6 . 6 \times 1 0 ^{- 34} \times 2 \times 1 0 ^{14} J}{1 . 6 \times 1 0 ^{- 19}} e V

∴ K_{m a x} = 0.83 e V