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What is the speed of an electron whose de Broglie wavelength is 0.1nm? By what potential difference, must have such an electron accelerated from an initial speed zero?

Last updated date: 24th Feb 2024
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IVSAT 2024
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Hint: Try to use the equation of De Broglie which relates wavelength with the mass and velocity. After obtaining the velocity, substitute it in the limiting condition of stopping potential in order to obtain the voltage.

Complete Solution :
- In order to answer our question, we need to know about the particle nature of electromagnetic radiation and the photoelectric effect. According to classical mechanics, the energy is emitted or absorbed discontinuously. Therefore, the energy of any electromagnetic radiation is proportional to its intensity and independent of its frequency or wavelength. Thus, the radiation emitted by the body being heated should have the same colour throughout, although, the intensity of the colour might change with temperature. Energy is transferred in packets. Each such packet is called a quantum. In case of light the quantum of energy is called a photon. The energy of the radiation is proportional to its frequency ($\upsilon $) and is expressed by the equation
 & E=h\upsilon \\
 & E=\dfrac{hc}{\lambda } \\

- E is the energy and h is Planck’s constant $6.626\times {{10}^{-34}}Js$. In the photoelectric effect, conservation of energy takes place. The energy which is supplied by the photon to the electron is used up to eject the electron from the metal atom and the remaining is used to eject the electron with maximum energy. The minimum potential applied by which velocity of ejected electrons becomes zero is known as stopping potential.
Here, e = $1.6\times {{10}^{-19}}C$. Now, let us come to our question. We can also write the De-Broglie’s equation as:
\[\lambda =\dfrac{h}{mv}\]

- Now, substituting the values from the question, we have:
  & {{10}^{-10}}=\dfrac{6.626\times {{10}^{-34}}}{9.1\times {{10}^{-31}}\times v} \\
 & So,\,v=7.28\times {{10}^{6}} \\
Now using the value of v, we have
 & \dfrac{1}{2}m{{v}^{2}}=eV \\
 & \dfrac{1}{2}\times \dfrac{9.1\times {{10}^{-31}}\times 52.99\times {{10}^{12}}}{1.6\times {{10}^{-19}}}=v \\
 & v=150V \\
So, we get the speed as $7.28\times {{10}^{6}}m/s$ and the potential difference to be 150 volts.

Note: When a photon having sufficient energy strikes an electron on the metal atom, it transfers energy instantaneously to the electron during the collision and electrons get ejected without any delay or time lag.
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