Question

The photoelectric work function for photo metal is 2.4eV . Among the four wavelengths, the wavelength of light for which photoemission does not take place is

• A. 200 nm
• B. 300 nm
• C. 700 nm
• D. 400 nm

Answer 1

The Correct Option is C

The photoelectric work function of the metal is given as \( 2.4 \, \text{eV} \). The energy of a photon is related to its wavelength by the equation: \[ E = \frac{hc}{\lambda} \] where:
\( E \) is the energy of the photon in electron volts (eV), 
\( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \)),
\( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
\( \lambda \) is the wavelength in meters. The photoemission will not take place unless the energy of the incident photon is greater than or equal to the work function of the metal. The energy corresponding to each wavelength is calculated using the above equation. For each wavelength, we compare the photon energy to the work function of \( 2.4 \, \text{eV} \).
For \( \lambda = 200 \, \text{nm} \), \[ E = \frac{(6.626 \times 10^{-34}) (3 \times 10^8)}{200 \times 10^{-9}} = 6.626 \, \text{eV} > 2.4 \, \text{eV} \] Photon energy is greater than the work function, so photoemission occurs.
For \( \lambda = 300 \, \text{nm} \), \[ E = 4.14 \, \text{eV} > 2.4 \, \text{eV} \] Photon energy is greater than the work function, so photoemission occurs.
For \( \lambda = 400 \, \text{nm} \), \[ E = 3.1 \, \text{eV} > 2.4 \, \text{eV} \] Photon energy is greater than the work function, so photoemission occurs.
For \( \lambda = 700 \, \text{nm} \), \[ E = 1.77 \, \text{eV} < 2.4 \, \text{eV} \] Photon energy is less than the work function, so photoemission does not occur. Thus, the wavelength for which photoemission does not take place is \( 700 \, \text{nm} \). Thus, the solution is \( 700 \, \text{nm} \).

Answer 2

The photoelectric effect occurs when photons of light strike a metal surface and eject electrons from it. The condition for photoemission is that the energy of the photons should be greater than or equal to the work function of the metal. The energy of a photon is given by: \[ E = \frac{hc}{\lambda} \] where \( h \) is Planck’s constant \(6.626 \times 10^{-34} \, \text{J.s}\) \(c\) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)), and \( \lambda \) is the wavelength of the light. The work function \( \phi \) of the metal is given as 2.4 eV. We can convert this energy to joules: \[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}, \quad \therefore 2.4 \, \text{eV} = 2.4 \times 1.6 \times 10^{-19} = 3.84 \times 10^{-19} \, \text{J} \] Now, to find the wavelength for which photo-emission does not occur, we will compare the energies of the different wavelengths to the work function. 1. For \( \lambda = 200 \, \text{nm} \): \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{200 \times 10^{-9}} = 9.94 \times 10^{-19} \, \text{J} \] Since \( E = 9.94 \times 10^{-19} \, \text{J} > 3.84 \times 10^{-19} \, \text{J} \), photo-emission will take place. 2. For \( \lambda = 300 \, \text{nm} \): \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} = 6.62 \times 10^{-19} \, \text{J} \] Since \( E = 6.62 \times 10^{-19} \, \text{J} > 3.84 \times 10^{-19} \, \text{J} \), photo-emission will take place. 3. For \( \lambda = 400 \, \text{nm} \): \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} = 4.97 \times 10^{-19} \, \text{J} \] Since \( E = 4.97 \times 10^{-19} \, \text{J} > 3.84 \times 10^{-19} \, \text{J} \), photo-emission will take place. 4. For \( \lambda = 700 \, \text{nm} \): \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{700 \times 10^{-9}} = 2.85 \times 10^{-19} \, \text{J} \] Since \( E = 2.85 \times 10^{-19} \, \text{J} < 3.84 \times 10^{-19} \, \text{J} \), photo-emission will not take place. Thus, the wavelength for which photo-emission does not take place is \( 700 \, \text{nm} \).