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Question

(C) The maximum kinetic energy for the photoelectrons is given by Einstein's photoelectric equation: $K_{\max} = h
u - \phi$,where $
u$ is the frequ

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become more than doubled

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(C) The maximum kinetic energy for the photoelectrons is given by Einstein's photoelectric equation: $K_{\max} = h
u - \phi$,where $
u$ is the frequency of incident light and $\phi$ is the work function of the metal.Since $K_{\max} = eV_0$,we have $eV_0 = h
u - \phi$  --- $(1)$When the frequency is doubled $(
u' = 2
u)$,the new stopping potential $V_0'$ is given by:$eV_0' = h(2
u) - \phi = 2h
u - \phi$We can rewrite this as:$eV_0' = 2(h
u - \frac{\phi}{2})$ --- $(2)$Comparing $(1)$ and $(2)$,since $h
u - \frac{\phi}{2} > h
u - \phi$,it follows that $eV_0' > 2eV_0$. Therefore,the stopping potential becomes more than double.

If the frequency of light in a photoelectric experiment is doubled,the stopping potential will

Similar Questions

The figure shows the plot of stopping potential $(V_0)$ versus $(1/\lambda)$ for three different metals. If $\phi$ is the work function,then which of the following is correct?

Light of wavelength $4000 \,\mathring A$ falls on a photosensitive metal and a negative $2 \,V$ potential stops the emitted electrons. The work function of the material (in $eV$) is approximately $(h = 6.6 \times 10^{-34} \,Js, \,e = 1.6 \times 10^{-19} \,C, \,c = 3 \times 10^8 \,m/s)$.

In a photoelectric experiment, the wavelength of the light incident on the metal is changed from $200 \, nm$ to $400 \, nm$. The decrease in the stopping potential is close to [Use $hc = 1240 \, eV \cdot nm$ where $h$ is Planck's constant and $c$ is the velocity of light]. (in $ \, V$)

When the wavelength of radiation falling on a metal is changed from $500 \, nm$ to $200 \, nm$,the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to $..... \, eV$.

The work function of a metallic surface is $5.01\ eV$. Photoelectrons are emitted when light of wavelength $2000\ \mathring{A}$ falls on it. The minimum potential difference required to stop the fastest photoelectrons is ................. $V$.

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