An electromagnetic wave of wavelength \(lambda\) is incident on a photosensitive surface of negligible work function. If \(m\) is mass of photoelectron emitted from the surface has de-Broglie wavelength \(lambda_d\), then
1. \(\lambda = \left(\frac{2h}{mc}\right)\lambda_d^2\)
2. \(\lambda = \left(\frac{2m}{hc}\right)\lambda_d^2\)
3. \(\lambda_d = \left(\frac{2mc}{h}\right)\lambda^2\)
4. \(\lambda = \left(\frac{2mc}{h}\right)\lambda_d^2\)
View Answer
With a negligible work function, the maximum kinetic energy of the emitted photoelectron is \(E = \frac{hc}{\lambda}\) and its de-Broglie wavelength is \(\lambda_d = \frac{h}{\sqrt{2mE}}\). Substituting \(E\) gives \(\lambda_d^2 = \frac{h\lambda}{2mc}\), which simplifies to \(\lambda = \left(\frac{2mc}{h}\right)\lambda_d^2\).
The potential difference that must be applied to stop the fastest moving photoelectrons emitted by a metal surface, having work function \(4.5\text{ eV}\), when ultraviolet light of \(2000A^\circ\) falls on it, will be
1. -0.7 V
2. -1.7 V
3. -1.2 V
4. -0.8 V
View Answer
Formula: \(eV_0 = E - \Phi_0\), where \(E = \frac{hc}{\lambda} = \frac{12400}{2000} = 6.2\text{ eV}\). Thus, the stopping potential \(V_0 = 6.2 - 4.5 = 1.7\text{ V}\), requiring an applied potential of \(-1.7\text{ V}\).
The stopping potential in the photoelectric experiment is \( 1.6\text{ V} \). The maximum kinetic energy of photoelectrons emitted is
1. \( 2.4 \times 10^{-19}\text{ J} \)
2. \( 2.56 \times 10^{-19}\text{ J} \)
3. \( 1.86 \times 10^{-19}\text{ J} \)
4. \( 1.4 \times 10^{-19}\text{ J} \)
View Answer
The maximum kinetic energy of photoelectrons is given by \( K_{\text{max}} = e V_0 \). Substituting the values: \( K_{\text{max}} = 1.6 \times 10^{-19}\text{ C} \times 1.6\text{ V} = 2.56 \times 10^{-19}\text{ J} \).
Assertion (A): Two photons having equal wavelengths have equal linear momentum.
Reason (R): When monochromatic light shows its photon character, each photon has a linear momentum \(p = \frac{h}{\lambda}\).
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
The linear momentum of a photon is given by \(p = h/\lambda\). If two photons have equal wavelengths \(\lambda\), then their linear momenta \(p\) must also be equal. The Reason (R) correctly states the formula and explains Assertion (A).
Assertion (A): A photon and an electron both have energy \(50\text{ eV}\). Both have different wavelengths.
Reason (R): Wavelength depends on energy and not on mass.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
For a photon, wavelength is \(\lambda_p = hc/E\). For an electron, de Broglie wavelength is \(\lambda_e = h/\sqrt{2mE}\). Since their formulas are different and \(\lambda_e\) depends on mass \(m\), their wavelengths will be different for the same energy. So (A) is true. Reason (R) is false because an electron's de Broglie wavelength depends on its mass.