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Two radiations of photon energies 1 \mathrm{eV} and 2.5 \mathrm{eV}, successively illuminate a photosensitive metallic surface of work function 0.5 \mathrm{eV}. The ratio of the maximum speeds of the emitted electrons is:


 

Option: 1

1:4


Option: 2

1:2


Option: 3

1:1


Option: 4

1:5


Answers (1)

best_answer

According to Einstein's photoelectric equation \frac{1}{2} \mathrm{mv}_{\max }^2=\mathrm{h\nu}-\phi_0

Where \frac{1}{2} \mathrm{mv}_{\max }^2 is the maximum kinetic energy of the emitted electrons,  h\nu is the incident energy and  \phi_0 is the work function of the metal.

\begin{aligned} & \therefore \frac{1}{2} \mathrm{mv}_{\max _1}^2=1 \mathrm{eV}-0.5 \mathrm{eV}=0.5 \mathrm{eV} \quad \quad \quad (i)\\ & \text { and } \frac{1}{2} \mathrm{mv}_{\max _2}^2=2.5 \mathrm{eV}-0.5 \mathrm{eV}=2 \mathrm{eV} \quad \quad (ii) \end{aligned}

Divide (i) and (ii), we get  

\frac{\mathrm{v}_{\max _1}^2}{\mathrm{v}_{\max _2}^2}=\frac{0.5}{2} ; \frac{\mathrm{v}_{\max _1}}{\mathrm{v}_{\max _2}}=\sqrt{\frac{0.5}{2}}=\frac{1}{2}
 

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Rishi

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