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Light of wavelength ′$\lambda$′ strikes a photosensitive surface and electrons are ejected with kinetic energy E. If the kinetic energy is to be increased to 2E, the wavelength must be changed to $\lambda '$ whereA. $\lambda ' = \dfrac{\lambda }{2}$B. $\lambda ' = 2\lambda$C. $\dfrac{\lambda }{2} < \lambda '$D. $\lambda ' > \lambda$

Last updated date: 13th Aug 2024
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Hint: When the photon is incident on metal then it contains energy which is able to overcome the attractive force with which the valence electron is bound to the shell of the atom of the metal. If the energy of the photon exceeds the minimum energy needed to eject the electron then the rest of the energy is transferred as kinetic energy of the ejected electrons.

Formula used:
$K = \dfrac{{hc}}{\lambda } - \phi$
where K is the kinetic energy of the emitted electron, h is the planck's constant, c is the speed of light, $\lambda$ is the wavelength of the photon and $\phi$ is the work function of the metal.

Complete step by step solution:
The threshold wavelength of the metal is the wavelength corresponding to the minimum energy which is sufficient to overcome the attractive force which causes the valence electron to be bound to the shell of the atom of the metal. As the energy of the photon is inversely proportional to the wavelength, so for a minimum value of the energy, the wavelength should be the maximum allowed wavelength.

The kinetic energy of the ejected electron is E when the wavelength of the light is $\lambda$. Using the energy formula for the emitted electron,
$E = \dfrac{{hc}}{\lambda } - \phi \ldots \ldots \left( i \right)$
We need to increase the kinetic energy of the ejected electron to 2E. Let the required wavelength is $\lambda '$. Then using the energy formula for the ejected electron, we get
$2E = \dfrac{{hc}}{{\lambda '}} - \phi$
From equation (i),
$2\left( {\dfrac{{hc}}{\lambda } - \phi } \right) = \dfrac{{hc}}{{\lambda '}} - \phi$

On simplifying, we get
$\dfrac{{2hc}}{\lambda } - 2\phi = \dfrac{{hc}}{{\lambda '}} - \phi \\$
$\Rightarrow \dfrac{{2hc}}{\lambda } - \dfrac{{hc}}{{\lambda '}} = \phi \\$
$\Rightarrow \left( {\dfrac{2}{\lambda } - \dfrac{1}{{\lambda '}}} \right) = \dfrac{\phi }{{hc}}$
The right hand side is positive, so
$\dfrac{2}{\lambda } - \dfrac{1}{{\lambda '}} > 0$
$\Rightarrow \dfrac{2}{\lambda } > \dfrac{1}{{\lambda '}} \\$
$\therefore \dfrac{\lambda }{2} < \lambda '$
Hence, the required wavelength of the light must be greater than half of the initial wavelength of the light to increase the energy of the ejected electron twice the initial.

Therefore, the correct option is C.

Note: As we know that the frequency is inversely proportional to the wavelength. So for threshold frequency, the minimum value of the frequency of the photon contains sufficient energy to eject the electron from the metal.