When UV light of wavelength 300 nm is incident on the metal surface having work function 2.13 eV, electron emission takes place. The stopping potential is : (Given hc = 1240 eV nm)
- 4 V
- 4.1 V
- 2 V
- 1.5 V
The Correct Option is C
Approach Solution - 1
The energy of the incident photon is given by:
\[\frac{hc}{\lambda} = eV_s + \phi\]
Substitute the given values:
\[\frac{1240}{300} \, \text{eV} - 2.13 \, \text{eV} = eV_s\]
\[4.13 \, \text{eV} - 2.13 \, \text{eV} = eV_s\]
\[V_s = 2 \, \text{V}\]
Approach Solution -2
To find the stopping potential when UV light of wavelength 300 nm is incident on a metal surface with a work function of 2.13 eV, we can use the photoelectric effect equation. According to the photoelectric equation:
\(E_k = h f - \Phi\)
Where:
- \(E_k\) is the kinetic energy of the emitted electrons.
- \(h\) is Planck's constant.
- \(f\) is the frequency of the incident light.
- \(\Phi\) is the work function of the metal.
The stopping potential \(V_0\) is related to the maximum kinetic energy by:
\(E_k = e V_0\)
Where \(e\) is the charge of an electron (approximately \(1.6 \times 10^{-19}\) coulombs).
First, convert the wavelength to energy using the relation:
\(E = \frac{hc}{\lambda}\)
Given \(hc = 1240 \text{ eV nm}\) and \(\lambda = 300 \text{ nm}\), we have:
\(E = \frac{1240 \text{ eV nm}}{300 \text{ nm}} = 4.13 \text{ eV}\)
Now, substitute this into the photoelectric equation:
\(E_k = E - \Phi = 4.13 \text{ eV} - 2.13 \text{ eV} = 2.00 \text{ eV}\)
Hence, the stopping potential \(V_0\) can be found using:
\(E_k = e V_0\)
\(V_0 = 2.00 \text{ volts}\)
Thus, the stopping potential is 2 V.
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