Photocell is illuminated by a point source of light, which is placed at a distance \(d\) from the cell. If the distance becomes \(2d\), then number of electrons emitted per second will be
1. Remain same
2. Four times
3. Two times
4. One-fourth
View Answer
Intensity of light from a point source is inversely proportional to the square of the distance, \(I \propto \frac{1}{d^2}\). Since the number of photoelectrons emitted per second is proportional to intensity, doubling the distance reduces the emission to one-fourth.
The number of photons per second on an average emitted by the source of monochromatic light of wavelength \(600\text{ nm}\), when it delivers the power of \(3.3 \times 10^{-3}\text{ watt}\) will be (\(h = 6.6 \times 10^{-34}\text{ J s}\))
1. \(10^{15}\)
2. \(10^{18}\)
3. \(10^{17}\)
4. \(10^{16}\)
View Answer
Power \(P = n \frac{hc}{\lambda}\), where \(n\) is the number of photons emitted per second. Substituting the given values, we get \(n = \frac{3.3 \times 10^{-3} \times 600 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} = 10^{16}\text{ s}^{-1}\).
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: \( K_{\text{max}} = 1.6 \times 10^{-19}\text{ C} \times 1.6\text{ V} = 2.56 \times 10^{-19}\text{ J} \).