Light of wavelength 330 nm falling on a piece of metal ejects electrons with sufficient energy which requires voltage V 0 to prevent them from reaching a collector. In the same setup, light of wavel

Question

Light of wavelength 330 nm falling on a piece of metal ejects electrons with sufficient energy which requires voltage $V_{0}$ to prevent them from reaching a collector. In the same setup, light of wavelength 220 nm, ejects electrons which require twice the voltage $V_{0}$ to stop them in reaching a collector. The numerical value of voltage $V_{0}$ is nearly

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Answer:

C

Solution:

$e V_{0} = \frac{h c}{330 nm} - ϕ 2 e V_{0} = \frac{h c}{220 nm} - ϕ V_{0} = 1240 (\frac{1}{220} - \frac{1}{330}) ≃ 1.88 V ≃ \frac{15}{8} V$

TestHub · Accepted Answer

​ e V 0 ​ = 330 nm h c ​ − ϕ ​ 2 e V 0 ​ = 220 nm h c ​ − ϕ ​ V 0 ​ = 1240 ( 220 1 ​ − 330 1 ​ ) ≃ 1.88 V ≃ 8 15 ​ V ​

Physics - Modern Physics Question with Solution | TestHub

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