In a photoelectric experiment,the wavelength | vedclass

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(C) According to Einstein's photoelectric equation:$\frac{hc}{\lambda_{1}} = \phi + eV_{1}$  ...... $(i)$$\frac{hc}{\lambda_{2}} = \phi + eV_{2}$  ...

Vedclass · Accepted Answer

$1$

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(C) According to Einstein's photoelectric equation:$\frac{hc}{\lambda_{1}} = \phi + eV_{1}$  ...... $(i)$$\frac{hc}{\lambda_{2}} = \phi + eV_{2}$  ...... $(ii)$Subtracting equation $(ii)$ from $(i)$:$\frac{hc}{\lambda_{1}} - \frac{hc}{\lambda_{2}} = e(V_{1} - V_{2})$$hc \left( \frac{1}{\lambda_{1}} - \frac{1}{\lambda_{2}} ight) = e \Delta V$$\Delta V = \frac{hc}{e} \left( \frac{\lambda_{2} - \lambda_{1}}{\lambda_{1} \lambda_{2}} ight)$Given $\frac{hc}{e} = 1240\, nm \cdot V$,$\lambda_{1} = 300\, nm$,and $\lambda_{2} = 400\, nm$:$\Delta V = 1240 	imes \left( \frac{400 - 300}{300 	imes 400} ight)$$\Delta V = 1240 	imes \left( \frac{100}{120000} ight)$$\Delta V = \frac{1240}{1200} \approx 1.03\, V$Thus,the decrease in stopping potential is close to $1\, V$.

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 ................ $V$ $\left( \frac{hc}{e} = 1240\, nm \cdot V \right)$

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