Modern Physics - NEET Physics Chapterwise MCQs & PYQs
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NEET Modern Physics MCQs & PYQs
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Question 201:
easy
Monochromatic 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.
The 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.
The 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.
Monochromatic 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 $n = \frac{P}{E} = \frac{2 \times 10^{-3}}{39.6 \times 10^{-20}} \approx 0.05 \times 10^{17} = 5 \times 10^{15}$.
11. 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 $\frac{1}{2}mv^2 = h\nu \implies v^2 = \frac{2h\nu}{m} \implies v = \sqrt{\frac{2h\nu}{m}}$.
12. 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 $v_1/v_2 = \sqrt{K_1/K_2} = \sqrt{1/4} = 1/2$.
13. A 200 W sodium street lamp emits yellow light of wavelength $0.6 \mu m$. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is. (2012 Pre)
Useful power for light emission is $P = 25\% \text{ of } 200 W = 50 W$. The energy of one photon is $E = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{0.6 \times 10^{-6}} = 3.3 \times 10^{-19} J$. The number of photons emitted per second is $n = \frac{P}{E} = \frac{50}{3.3 \times 10^{-19}} \approx 1.5 \times 10^{20}$.
14. Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the materials is: (2012 Pre)
Energy of incident radiation $E = 13.6 (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 \times \frac{3}{4} = 10.2 eV$. The stopping potential is $3.57 V$, so $K_{max} = 3.57 eV$. Work function $W = E - K_{max} = 10.2 - 3.57 = 6.63 eV$. Using $W = h\nu_0$, we get threshold frequency $\nu_0 = \frac{6.63 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 1.6 \times 10^{15} Hz$.
15. The threshold frequency for a photosensitive metal is $3.3 \times 10^{14} Hz$. If light of frequency $8.2 \times 10^{14} Hz$ is incident on this metal, the cut-off voltage for the photoelectric emission is nearly: (2011 Mains)
16. In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is: (2011 Pre)
The stopping potential $V_0$ is numerically equal to the maximum kinetic energy of the emitted photoelectrons expressed in electron-volts (eV). Since $K_{max} = 0.5 eV$, the stopping potential is $0.5 V$.