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  • The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectron gets:  Option

The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectron gets:
 

Option: 1

doubled the earlier value
 


Option: 2

unchanged
 


Option: 3

more than doubled

 


Option: 4

 less than doubled


Answers (1)

best_answer

Let K_1 and K_2 be the kinetic energy of photoelectrons for incident light of frequency \mathrm{v} and 2 \mathrm{v} respectively.
According to Einstein's photoelectric equation
\begin{aligned} & \mathrm{hv}=\mathrm{K}_1+\phi_0 \quad \quad \quad \dots (i) \\ & \text {and } \mathrm{h} 2 \mathrm{v}=\mathrm{K}_2+\phi_0 \quad \dots (ii) \\ & \therefore 2\left(\mathrm{~K}_1+\phi_0\right)=\mathrm{K}_2+\phi_0 \text { or } \mathrm{K}_2=2 \mathrm{~K}_1+\phi_0 \end{aligned}
It means the kinetic energy of the photoelectrons becomes more than doubled.

Posted by

Deependra Verma

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