Question:
Radiation, with wavelength 6561 $?$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} T$. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
Radiation, with wavelength 6561 $?$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} T$. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
Updated On: Jan 18, 2023
- 0.8 eV
- 1.1 eV
- 1.8 eV
- 1.6 eV
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The Correct Option is A
Solution and Explanation
Let the work function be $\phi.$
$\therefore KE_{max}=\frac{hc}{\lambda}-\phi$
Again, $R_{max}=\frac{\sqrt{2mKE_{max}}}{qB}=\frac{\sqrt{2m\left(\frac{hc}{\lambda}-\phi\right)}}{qB}$
$\therefore \frac{R^{2}_{max}q^{2}B^{2}}{2m}=\frac{hc}{\lambda}-\phi$
$\therefore \phi=\frac{hc}{\lambda}-\frac{R^{2}_{max}q^{2}B^{2}}{2m}=1.0899\,eV\,\approx\,1.1\,eV$
$\therefore KE_{max}=\frac{hc}{\lambda}-\phi$
Again, $R_{max}=\frac{\sqrt{2mKE_{max}}}{qB}=\frac{\sqrt{2m\left(\frac{hc}{\lambda}-\phi\right)}}{qB}$
$\therefore \frac{R^{2}_{max}q^{2}B^{2}}{2m}=\frac{hc}{\lambda}-\phi$
$\therefore \phi=\frac{hc}{\lambda}-\frac{R^{2}_{max}q^{2}B^{2}}{2m}=1.0899\,eV\,\approx\,1.1\,eV$
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Concepts Used:
Photoelectric Effect
When light shines on a metal, electrons can be ejected from the surface of the metal in a phenomenon known as the photoelectric effect. This process is also often referred to as photoemission, and the electrons that are ejected from the metal are called photoelectrons.
Photoelectric Effect Formula:
According to Einstein’s explanation of the photoelectric effect :
The energy of photon = energy needed to remove an electron + kinetic energy of the emitted electron
i.e. hν = W + E
Where,
- h is Planck’s constant.
- ν is the frequency of the incident photon.
- W is a work function.
- E is the maximum kinetic energy of ejected electrons: 1/2 mv².
Laws of Photoelectric Effect:
- The photoelectric current is in direct proportion to the intensity of light, for a light of any given frequency; (γ > γ Th).
- There exists a certain minimum (energy) frequency for a given material, called threshold frequency, below which the discharge of photoelectrons stops completely, irrespective of how high the intensity of incident light is.
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- The process of photo-emission is an instantaneous process.