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

 

  • Option 1)

    1 : 2

  • Option 2)

    1 : 1

  • Option 3)

    1 : 5

     

  • Option 4)

    1 : 4

 

Answers (1)

best_answer

\frac{1}{2}mv^{2}=\frac{hc}{\lambda }-\phi \\ Case\:I:\frac{1}{2}mv_{1}^{2}=1-0.5=0.5ev\rightarrow 1\\ Case II=\frac{1}{2}mv_{2}^{2}=(2.5-0.5)ev=2ev\rightarrow 2\\

\therefore\: Divide\: 1\: and\: 2

\frac{V_{1}^{2}}{V_{2}^{2}}=\frac{1}{4}\:or\:\frac{V_{1}}{V_{2}}=\frac{1}{2}


Option 1)

1 : 2

This solution is correct 

Option 2)

1 : 1

This solution is incorrect 

Option 3)

1 : 5

 

This solution is incorrect 

Option 4)

1 : 4

This solution is incorrect 

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prateek

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