Question:
If a radiation of energy $5.2$ eV falls on the photosensitive surfaces of Mo and Ni, they emit photoelectrons with maximum kinetic energy of $0.5$ eV and $1$ eV, respectively. Then the work function of
If a radiation of energy $5.2$ eV falls on the photosensitive surfaces of Mo and Ni, they emit photoelectrons with maximum kinetic energy of $0.5$ eV and $1$ eV, respectively. Then the work function of
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Photoelectric effect:
- $E = \phi + K_{\max}$
- Work function = incident energy - kinetic energy
Updated On: Apr 30, 2026
- Mo is $2.6$ eV
- Ni is $6.2$ eV
- Mo is $6.2$ eV
- Ni is $4.2$ eV
- Mo is $4.2$ eV
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The Correct Option is D
Solution and Explanation
Concept:
Photoelectric equation:
\[
E = \phi + K_{\max}
\]
Step 1: For Ni.
\[ \phi_{\text{Ni}} = E - K_{\max} = 5.2 - 1 = 4.2\ \text{eV} \]
Step 2: For Mo.
\[ \phi_{\text{Mo}} = 5.2 - 0.5 = 4.7\ \text{eV} \]
Step 3: Check options.
Only correct statement given: \[ \phi_{\text{Ni}} = 4.2\ \text{eV} \]
Step 1: For Ni.
\[ \phi_{\text{Ni}} = E - K_{\max} = 5.2 - 1 = 4.2\ \text{eV} \]
Step 2: For Mo.
\[ \phi_{\text{Mo}} = 5.2 - 0.5 = 4.7\ \text{eV} \]
Step 3: Check options.
Only correct statement given: \[ \phi_{\text{Ni}} = 4.2\ \text{eV} \]
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