
The work function of a certain metal is $2.3eV$. If the light of wavenumber $2 \times {10^6}{m^{ - 1}}$ falls on it, the kinetic energies of the fastest and slowest ejected electron will be respectively:
A. $2.48eV,0.18eV$
B. $0.18eV,Zero$
C. $2.30eV,Zero$
D. $0.18eV,0.18eV$
Answer
495.3k+ views
Hint: Photoelectric effect is the process in which electromagnetic radiation is made to fall on the metal surface due to which emission of electrons takes place. The part of the energy of the incident light is used in doing work against the work function of the metal to remove an electron from it and the remaining goes to its kinetic energy. The higher the energy of radiation, the higher will be the kinetic energy of the ejected electron.
Complete step by step solution:
Let us write the information given in the question.
The work function of the metal $\phi = 2.3eV$, wave number of lightwave $\bar \nu = \dfrac{1}{\lambda } = 2 \times {10^6}{m^{ - 1}}$
We have to calculate the kinetic energy of the fastest and slowest electron.
The relation between kinetic energy and work function is given below.
$KE = h\nu - \phi = \dfrac{{hc}}{\lambda } - \phi $
Here, $h$ is Planck’s constant, $\nu $is frequency, $\lambda $ is the wavelength, $c$ is the speed of light, and $\phi $ is the work function of the metal.
Let us put the values in the above expression, this will give the maximum kinetic energy.
${(KE)_{max}} = \dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8} \times 2 \times {{10}^6}}}{{1.6 \times {{10}^{ - 19}}}} - 2.3 = 2.475 - 2.3 = 0.175eV$
So the fastest electron will be ejected with kinetic energy $0.175eV$.
Now, the slowest electron will be the one that uses all the energy of the photon to overcome the collision. So, the smallest possible kinetic energy will be zero.
${\left( {KE} \right)_{\min }} = 0$
Therefore, the kinetic energy of the fastest electron is $0.175eV$ and the kinetic energy of the slowest electron is zero.
Hence, the correct option is (B) $0.18eV,0eV$.
Note:
Work function is the minimum energy required to remove the electron from that particular metal, it is the characteristic property of the metal.For the same frequency of light, the kinetic energy of ejected electrons will be different for different metals. Theone having less work function will eject the electrons of higher kinetic energy.
Speed of light is $3 \times {10^8}m/s$, in a vacuum and $h$is Planck’s constant and its value is $6.6 \times {10^{ - 34}}J - s$.
Complete step by step solution:
Let us write the information given in the question.
The work function of the metal $\phi = 2.3eV$, wave number of lightwave $\bar \nu = \dfrac{1}{\lambda } = 2 \times {10^6}{m^{ - 1}}$
We have to calculate the kinetic energy of the fastest and slowest electron.
The relation between kinetic energy and work function is given below.
$KE = h\nu - \phi = \dfrac{{hc}}{\lambda } - \phi $
Here, $h$ is Planck’s constant, $\nu $is frequency, $\lambda $ is the wavelength, $c$ is the speed of light, and $\phi $ is the work function of the metal.
Let us put the values in the above expression, this will give the maximum kinetic energy.
${(KE)_{max}} = \dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8} \times 2 \times {{10}^6}}}{{1.6 \times {{10}^{ - 19}}}} - 2.3 = 2.475 - 2.3 = 0.175eV$
So the fastest electron will be ejected with kinetic energy $0.175eV$.
Now, the slowest electron will be the one that uses all the energy of the photon to overcome the collision. So, the smallest possible kinetic energy will be zero.
${\left( {KE} \right)_{\min }} = 0$
Therefore, the kinetic energy of the fastest electron is $0.175eV$ and the kinetic energy of the slowest electron is zero.
Hence, the correct option is (B) $0.18eV,0eV$.
Note:
Work function is the minimum energy required to remove the electron from that particular metal, it is the characteristic property of the metal.For the same frequency of light, the kinetic energy of ejected electrons will be different for different metals. Theone having less work function will eject the electrons of higher kinetic energy.
Speed of light is $3 \times {10^8}m/s$, in a vacuum and $h$is Planck’s constant and its value is $6.6 \times {10^{ - 34}}J - s$.
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