CBSE Class 12 Physics Chapter 21: NCERT Solutions

The potential difference through which the electrons are accelerated is given as $V\; =\; 30\; \backslash text\{\; kV\}\; =\; 3\; \backslash times\; 10^4\; \backslash text\{\; V\}$. This means the maximum kinetic energy an electron can possess is equal to the energy gained by moving through this potential difference.

The energy of the electrons is $E\; =\; eV$, where $e$ is the elementary charge ($1.6\; \backslash times\; 10^\{-19\}\; \backslash text\{\; C\}$). Therefore, the maximum energy of the electrons is $E\; =\; (1.6\; \backslash times\; 10^\{-19\}\; \backslash text\{\; C\})\; \backslash times\; (3\; \backslash times\; 10^4\; \backslash text\{\; V\})\; =\; 4.8\; \backslash times\; 10^\{-15\}\; \backslash text\{\; J\}$. This energy can also be expressed in electron volts as $E\; =\; 30\; \backslash text\{\; keV\}$ or $3\; \backslash times\; 10^4\; \backslash text\{\; eV\}$.

1. Maximum frequency of X-rays:

   When an electron loses its entire kinetic energy in a single collision, it produces an X-ray photon with the maximum possible frequency (

   u_{max}). The energy of this photon is equal to the maximum kinetic energy of the electron.

   Using the relation E = h

   u_{max}, where h is Planck's constant (6.626 \times 10^{-34} \text{ Js}), we can find the maximum frequency:

   u_{max} = \frac{E}{h} = \frac{4.8 \times 10^{-15} \text{ J}}{6.626 \times 10^{-34} \text{ Js}} \approx 7.24 \times 10^{18} \text{ Hz}

   Hence, the maximum frequency of X-rays produced is approximately 7.24 \times 10^{18} \text{ Hz}.

2. Minimum wavelength of X-rays:

   The minimum wavelength (\lambda_{min}) corresponds to the maximum frequency (

   u_{max}) of the X-rays, as wavelength and frequency are inversely related by the speed of light (c).

   The relation is \lambda_{min} = \frac{c}{

   u_{max}}, where c = 3 \times 10^8 \text{ m/s}.

   \lambda_{min} = \frac{3 \times 10^8 \text{ m/s}}{7.24 \times 10^{18} \text{ Hz}} \approx 4.14 \times 10^{-11} \text{ m}

   This can also be expressed in nanometers: 4.14 \times 10^{-11} \text{ m} = 0.0414 \text{ nm}.

   Hence, the minimum wavelength of X-rays produced is approximately 4.14 \times 10^{-11} \text{ m} or 0.0414 \text{ nm}.

Given: Work function of caesium metal, $\backslash phi\_0\; =\; 2.14\; \backslash text\{\; eV\}$ Frequency of incident light, $u\; =\; 6.0\; \backslash times\; 10^\{14\}\; \backslash text\{\; Hz\}$ Planck's constant, $h\; =\; 6.626\; \backslash times\; 10^\{-34\}\; \backslash text\{\; Js\}$ Charge of an electron, $e\; =\; 1.6\; \backslash times\; 10^\{-19\}\; \backslash text\{\; C\}$

1. Maximum kinetic energy of the emitted electrons:

   According to Einstein's photoelectric equation, the maximum kinetic energy (K_{max}) of the emitted photoelectrons is given by the energy of the incident photon minus the work function of the metal.

   K_{max} = h

   u - \phi_0

   First, calculate the energy of the incident photon in Joules: E_{photon} = h

   u = (6.626 \times 10^{-34} \text{ Js}) \times (6.0 \times 10^{14} \text{ Hz}) = 3.9756 \times 10^{-19} \text{ J}.

   Convert the photon energy to electron volts: E_{photon} = \frac{3.9756 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \approx 2.485 \text{ eV}.

   Now, calculate the maximum kinetic energy:

   K_{max} = E_{photon} - \phi_0 = 2.485 \text{ eV} - 2.14 \text{ eV} = 0.345 \text{ eV}

   Hence, the maximum kinetic energy of the emitted electrons is 0.345 \text{ eV}.

2. Stopping potential:

   The stopping potential ($V\_0$) is the minimum reverse potential difference required to stop the most energetic photoelectrons from reaching the collector. The work done by the stopping potential on an electron is equal to its maximum kinetic energy.

   $K\_\{max\}\; =\; eV\_0$

   Therefore, $V\_0\; =\; \backslash frac\{K\_\{max\}\}\{e\}$.

   Using the kinetic energy in Joules: $K\_\{max\}\; =\; 0.345\; \backslash text\{\; eV\}\; \backslash times\; 1.6\; \backslash times\; 10^\{-19\}\; \backslash text\{\; J/eV\}\; =\; 0.552\; \backslash times\; 10^\{-19\}\; \backslash text\{\; J\}$.

   $$V\_0\; =\; \backslash frac\{0.552\; \backslash times\; 10^\{-19\}\; \backslash text\{\; J\}\}\{1.6\; \backslash times\; 10^\{-19\}\; \backslash text\{\; C\}\}\; =\; 0.345\; \backslash text\{\; V\}$$

   Hence, the stopping potential is $0.345\; \backslash text\{\; V\}$.

3. Maximum speed of the emitted photoelectrons:

   The maximum kinetic energy of an electron is also given by $K\_\{max\}\; =\; \backslash frac\{1\}\{2\}mv\_\{max\}^2$, where $m$ is the mass of the electron and $v\_\{max\}$ is its maximum speed.

   The mass of an electron is $m\; =\; 9.1\; \backslash times\; 10^\{-31\}\; \backslash text\{\; kg\}$.

   Rearranging the formula to solve for $v\_\{max\}$:

   $v\_\{max\}\; =\; \backslash sqrt\{\backslash frac\{2K\_\{max\}\}\{m\}\}$

   $$v\_\{max\}\; =\; \backslash sqrt\{\backslash frac\{2\; \backslash times\; (0.552\; \backslash times\; 10^\{-19\}\; \backslash text\{\; J\})\}\{9.1\; \backslash times\; 10^\{-31\}\; \backslash text\{\; kg\}\}\}\; =\; \backslash sqrt\{\backslash frac\{1.104\; \backslash times\; 10^\{-19\}\}\{9.1\; \backslash times\; 10^\{-31\}\}\}\; =\; \backslash sqrt\{0.1213\; \backslash times\; 10^\{12\}\}\; \backslash approx\; \backslash sqrt\{1.213\; \backslash times\; 10^\{11\}\}\; \backslash approx\; 3.48\; \backslash times\; 10^5\; \backslash text\{\; m/s\}$$

   Hence, the maximum speed of the emitted photoelectrons is approximately $3.48\; \backslash times\; 10^5\; \backslash text\{\; m/s\}$.