JEE MAIN - Chemistry (2021 - 26th August Evening Shift - No. 16)
A metal surface is exposed to 500 nm radiation. The threshold frequency of the metal for photoelectric current is 4.3 $$\times$$ 1014 Hz. The velocity of ejected electron is ____________ $$\times$$ 105 ms$$-$$1 (Nearest integer)
[Use : h = 6.63 $$\times$$ 10$$-$$34 Js, me = 9.0 $$\times$$ 10$$-$$31 kg]
[Use : h = 6.63 $$\times$$ 10$$-$$34 Js, me = 9.0 $$\times$$ 10$$-$$31 kg]
Answer
5
Explanation
$$\upsilon $$ : speed of electron having max. K.E.
$$\Rightarrow$$ from Einstein equation : E = $$\phi$$ + K.E.max
$$ \Rightarrow {{hc} \over \lambda } = h{\upsilon _0} + {1 \over 2}m{v^2}$$
$$ \Rightarrow {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {500 \times {{10}^{ - 9}}}} = 6.63 \times {10^{ - 34}} \times 4.3 \times {10^{14}} + {1 \over 2}m{v^2}$$
$$ \Rightarrow {{6.63 \times 30 \times {{10}^{ - 20}}} \over 5} = 6.63 \times 4.3 \times {10^{ - 20}} + {1 \over 2}m{v^2}$$
$$ \Rightarrow 11.271 \times {10^{ - 20}}J = {1 \over 2} \times 9 \times {10^{ - 31}} \times {\upsilon ^2}$$
$$\Rightarrow$$ $$\upsilon $$ = 5 $$\times$$ 105 m/sec.
$$\Rightarrow$$ from Einstein equation : E = $$\phi$$ + K.E.max
$$ \Rightarrow {{hc} \over \lambda } = h{\upsilon _0} + {1 \over 2}m{v^2}$$
$$ \Rightarrow {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {500 \times {{10}^{ - 9}}}} = 6.63 \times {10^{ - 34}} \times 4.3 \times {10^{14}} + {1 \over 2}m{v^2}$$
$$ \Rightarrow {{6.63 \times 30 \times {{10}^{ - 20}}} \over 5} = 6.63 \times 4.3 \times {10^{ - 20}} + {1 \over 2}m{v^2}$$
$$ \Rightarrow 11.271 \times {10^{ - 20}}J = {1 \over 2} \times 9 \times {10^{ - 31}} \times {\upsilon ^2}$$
$$\Rightarrow$$ $$\upsilon $$ = 5 $$\times$$ 105 m/sec.
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