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  • Stuck here, help me understand: - In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 n - Dual Nature of Matter and Radiation - JEE Main

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to : \left ( \frac{hc}{e} = 1240 \ nm-V \right )

 

  • Option 1)

    1.0 V

  • Option 2)

    0.5 V

  • Option 3)

    1.5 V

  • Option 4)

    2.0 V

Answers (1)

best_answer

 

Stopping Potential /Cut-off Potential -

It is defined as the potential necessary to stop any electron from reaching the other side.

-

 

\lambda _{1}=300nm

\lambda _{2}=400nm

\frac{hc}{e}=1240nm-v

For 1 

\frac{hc}{\lambda _{1}}=\phi +ev_{1}\cdots \cdots (1)

For 2

\frac{hc}{\lambda _{2}}=\phi +ev_{2}\cdots \cdots (2)

(2) - (1)

\Rightarrow hC\left ( \frac{1}{\lambda _{2}}-\frac{1}{\lambda _{1}}\right )=e\left ( v_{2}-v_{1} \right )

\left ( \frac{hc}{e} \right )\left ( \frac{\lambda _{2}-\lambda _{1}}{\lambda _{2}\lambda _{1}} \right )=v_{2}-v_{1}

So v_{2}-v_{1}=\frac{\left ( 400-300 \right )}{400\times 300}\times 1240

                      =\frac{100}{400\times 300}\times 1240

v_{2}-v_{1}\approx 1v


Option 1)

1.0 V

Option 2)

0.5 V

Option 3)

1.5 V

Option 4)

2.0 V

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