Question 11:
easy18. Photoelectric emission occurs only when the incident light has more than a certain minimum: (2011 Pre)
Photoelectric emission takes place only when the frequency of the incident light is greater than a characteristic minimum frequency for the given metal, known as the threshold frequency.
Question 12:
easyWhen monochromatic radiation of intensity I falls on a metal surface, the number of photoelectron and their maximum kinetic energy are N and T respectively. If the intensity of radiation is 2I, the number of emitted electrons and their maximum kinetic energy are respectively:
(2010 Mains)
The number of emitted photoelectrons per second is directly proportional to the intensity of the incident radiation, so it becomes 2N. However, the maximum kinetic energy depends only on the frequency of the incident radiation and the material's work function, so it remains T.
Question 13:
easyThe potential difference that must be applied to stop the fastest photo electrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be:
(2010 Pre)
The energy of incident photons is $E = \frac{1240 eV nm}{\lambda(nm)} = \frac{1240}{200} = 6.2 eV$. The maximum kinetic energy of emitted photoelectrons is $K_{max} = E - W = 6.2 - 5.01 = 1.19 eV \approx 1.2 eV$. Therefore, the stopping potential difference magnitude is $1.2 V$ (Often referenced as $1.2 V$ depending on convention for potential difference).
Question 14:
easyA source $S_1$ is producing $10^{15}$ photons per second of wavelength $5000 \mathring{A}$. Another source $S_2$ is producing $1.02 \times 10^{15}$ photons per second of wavelength $5100 \mathring{A}$. Then (power of $S_2$)/(power of $S_1$) is equal to:
(2010 Pre)
Power is given by $P = \frac{n h c}{\lambda}$, where n is the number of photons emitted per second. Thus, $\frac{P_2}{P_1} = \frac{n_2 / \lambda_2}{n_1 / \lambda_1} = \left(\frac{n_2}{n_1}\right) \left(\frac{\lambda_1}{\lambda_2}\right) = \left(\frac{1.02 \times 10^{15}}{10^{15}}\right) \left(\frac{5000}{5100}\right) = 1.02 \times \frac{50}{51} = 1.00$.
Question 15:
easyMonochromatic light of wavelength 667 nm is produced by a helium neon laser. The power emitted is 9 mW. The number of photons arriving per sec. on the average at a target irradiated by this beam is:
(2009)
Number of photons per second is $n = \frac{P}{E} = \frac{P \lambda}{h c}$. Substituting the values, $n = \frac{9 \times 10^{-3} \times 667 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} = \frac{6003 \times 10^{-12}}{19.8 \times 10^{-26}} \approx 3 \times 10^{16}$ photons per second.
Question 16:
easyThe number of photo electrons emitted for light of a frequency $\nu$ (higher than the threshold frequency $\nu_0$) is proportional to:
(2009)
According to the laws of photoelectric emission, the number of photoelectrons emitted per second (photoelectric current) is directly proportional to the intensity of incident light, provided the frequency is above the threshold frequency.
Question 17:
easyThe work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the:
(2008)
The energy of incident radiation is $E = W + eV_0 = 6.2 eV + 5.0 eV = 11.2 eV$. The wavelength is $\lambda = \frac{1240 eV nm}{E(eV)} = \frac{1240}{11.2} \approx 110 nm$. This wavelength lies in the ultraviolet region of the electromagnetic spectrum.
Question 18:
easyMonochromatic light of frequency $6.0 \times 10^{14}$ Hz is produced by a laser. The power emitted is $2 \times 10^{-3}$ W. The number of photons emitted, on the average, by the sources per second is:
(2007)
Energy of a single photon is $E = h\nu = 6.6 \times 10^{-34} \times 6.0 \times 10^{14} = 39.6 \times 10^{-20} J$. The number of photons emitted per second is $n = \frac{P}{E} = \frac{2 \times 10^{-3}}{39.6 \times 10^{-20}} \approx 0.05 \times 10^{17} = 5 \times 10^{15}$.
Question 19:
easy11. For photoelectric emission from certain metal the cut-off frequency is $\nu$. If radiation of frequency $2\nu$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be: (m is the electron mass) (2013)
From Einstein's photoelectric equation, $K_{max} = h\nu_{incident} - h\nu_{threshold}$. Here, incident frequency is $2\nu$ and threshold frequency is $\nu$. So, $K_{max} = h(2\nu) - h\nu = h\nu$. Since $K_{max} = \frac{1}{2}mv^2$, we have $\frac{1}{2}mv^2 = h\nu \implies v^2 = \frac{2h\nu}{m} \implies v = \sqrt{\frac{2h\nu}{m}}$.
Question 20:
easy12. Two radiations of photons energies 1 eV and 2.5 eV, successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is: (2012 Mains)
Maximum kinetic energy is given by $K = E - W$. For the first radiation, $K_1 = 1.0 - 0.5 = 0.5 eV$. For the second radiation, $K_2 = 2.5 - 0.5 = 2.0 eV$. The ratio of their kinetic energies is $K_1/K_2 = 0.5/2.0 = 1/4$. Since $K \propto v^2$, the ratio of speeds is $v_1/v_2 = \sqrt{K_1/K_2} = \sqrt{1/4} = 1/2$.