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Step-by-Step Solution
Step 1: Understand the Given Data
• Work function of the material, $ \phi = 6.2 \,\text{eV}$
• Stopping potential, $V_{0} = 5\,\text{V}$
• Maximum kinetic energy of emitted electrons, $K_{\text{max}} = e\,V_{0} = 5\,\text{eV}$
Step 2: Relate Photon Energy to Work Function and Kinetic Energy
According to the photoelectric effect equation, the photon energy $h\nu$ must be equal to the sum of the work function and the maximum kinetic energy:
$$
h \nu = \phi + K_{\text{max}}
$$
Substituting the values:
$$
h \nu = 6.2\,\text{eV} + 5\,\text{eV} = 11.2\,\text{eV}
$$
Step 3: Calculate the Wavelength
The wavelength $ \lambda $ of the incident radiation is related to the photon energy $E$ (in eV) by the relation:
$$
\lambda = \frac{hc}{E},
$$
where $hc \approx 12400\,\text{eV·Å}$. Thus,
$$
\lambda = \frac{12400\,\text{eV·Å}}{11.2\,\text{eV}} = 1107\,\text{Å}
$$
Converting to nanometers,
$$
1\,\text{Å} = 10^{-1}\,\text{nm},
\quad
\lambda = 110.7\,\text{nm}.
$$
Step 4: Identify the Region of the Spectrum
A wavelength of approximately $110.7\,\text{nm}$ lies in the ultraviolet (UV) region of the electromagnetic spectrum.
Final Answer
The wavelength of the incident radiation for which the stopping potential is 5 V falls in the Ultraviolet region.
