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QJEEM21Ch62be: JEE Main 2021 Chemistry - Class 11 JEEM21Ch62be Solution: A metal surface is exposed to 500 nm radiation. The threshold…

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]

Step-by-Step Verified Solution

υ\upsilon : speed of electron having max. K.E.

⇒\Rightarrow from Einstein equation : E = ϕ\phi + K.E.max

⇒hcλ=hυ0+12mv2\Rightarrow {{hc} \over \lambda } = h{\upsilon _0} + {1 \over 2}m{v^2}

⇒6.63×10−34×3×108500×10−9=6.63×10−34×4.3×1014+12mv2\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}

⇒6.63×30×10−205=6.63×4.3×10−20+12mv2\Rightarrow {{6.63 \times 30 \times {{10}^{ - 20}}} \over 5} = 6.63 \times 4.3 \times {10^{ - 20}} + {1 \over 2}m{v^2}

⇒11.271×10−20J=12×9×10−31×υ2\Rightarrow 11.271 \times {10^{ - 20}}J = {1 \over 2} \times 9 \times {10^{ - 31}} \times {\upsilon ^2}

⇒\Rightarrow υ\upsilon = 5 ×\times 105 m/sec.

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