CBSE Class 12 Physics Chapter 11: Dual Nature of Radiation and Matter NCERT Solutions

Given the accelerating potential for the electrons, $V = 30 \text{ kV} = 3 \times 10^4 \text{ V}$.

The energy gained by each electron is equal to the potential difference multiplied by the elementary charge ($e$).

Energy of electrons, $E = eV = (1.6 \times 10^{-19} \text{ C}) \times (3 \times 10^4 \text{ V}) = 4.8 \times 10^{-15} \text{ J}$.

This energy can also be expressed in electron volts (eV): $E = 30 \text{ keV} = 30 \times 10^3 \text{ eV}$.

1. Maximum frequency of X-rays:

When an electron is suddenly stopped, its entire kinetic energy is converted into the energy of a single photon of an X-ray. The maximum frequency ($v_{max}$) of the X-rays produced is related to the electron's energy by the equation $E = hv_{max}$, where $h$ is Planck's constant ($6.626 \times 10^{-34} \text{ Js}$).

Therefore, $v_{max} = \frac{E}{h}$.

Substituting the values:

$$v\_\{max\}\; =\; \backslash frac\{4.8\; \backslash times\; 10^\{-15\}\; \backslash text\{\; J\}\}\{6.626\; \backslash times\; 10^\{-34\}\; \backslash text\{\; Js\}\}\; \backslash approx\; 7.24\; \backslash times\; 10^\{18\}\; \backslash text\{\; Hz\}$$

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

1. Minimum wavelength of X-rays:

The minimum wavelength ($\lambda_{min}$) of the X-rays corresponds to the maximum frequency. The relationship between wavelength, speed of light ($c$), and frequency ($v$) is $\lambda = \frac{c}{v}$.

Using the maximum frequency, we get the minimum wavelength:

$$\backslash lambda\_\{min\}\; =\; \backslash frac\{c\}\{v\_\{max\}\}\; =\; \backslash frac\{3\; \backslash times\; 10^8\; \backslash text\{\; m/s\}\}\{7.24\; \backslash times\; 10^\{18\}\; \backslash text\{\; Hz\}\}\; \backslash approx\; 4.14\; \backslash times\; 10^\{-11\}\; \backslash text\{\; m\}$$

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

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

Given:

• Work function of caesium metal, $\phi_0 = 2.14 \text{ eV}$.
• Frequency of incident light, $ u = 6.0 \times 10^{14} \text{ Hz}$.
• Planck's constant, $h = 6.626 \times 10^{-34} \text{ Js}$.
• Charge of an electron, $e = 1.6 \times 10^{-19} \text{ C}$.
• Mass of an electron, $m_e = 9.1 \times 10^{-31} \text{ kg}$.

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:

K_{max} = h

u - \phi_0

First, calculate the energy of the incident photons in Joules:

E_{photon} = h

u = (6.626 \times 10^{-34} \text{ Js}) \times (6 \times 10^{14} \text{ Hz}) = 3.9756 \times 10^{-19} \text{ J}

Convert the work function from eV to Joules:

\phi_0 = 2.14 \text{ eV} \times (1.6 \times 10^{-19} \text{ J/eV}) = 3.424 \times 10^{-19} \text{ J}

Now, calculate the maximum kinetic energy in Joules:

K_{max} = (3.9756 \times 10^{-19} \text{ J}) - (3.424 \times 10^{-19} \text{ J}) = 0.5516 \times 10^{-19} \text{ J}

Convert the kinetic energy back to electron volts (eV) for easier comparison with the work function:

K_{max} = \frac{0.5516 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \approx 0.345 \text{ eV}

Hence, the maximum kinetic energy of the emitted electrons is approximately 0.345 eV.

1. Stopping potential:

The stopping potential ($V_0$) is the potential difference required to stop the most energetic photoelectrons. The kinetic energy is related to the stopping potential by $K_{max} = eV_0$.

Using the maximum kinetic energy in Joules:

$$V\_0\; =\; \backslash frac\{K\_\{max\}\}\{e\}\; =\; \backslash frac\{0.5516\; \backslash times\; 10^\{-19\}\; \backslash text\{\; J\}\}\{1.6\; \backslash times\; 10^\{-19\}\; \backslash text\{\; C\}\}\; \backslash approx\; 0.345\; \backslash text\{\; V\}$$

Hence, the stopping potential is approximately 0.345 V.

1. Maximum speed of the emitted photoelectrons:

The maximum kinetic energy of an electron is also given by $K_{max} = \frac{1}{2}m_e v_{max}^2$, where $m_e$ is the mass of the electron and $v_{max}$ is its maximum speed.

Rearranging the formula to solve for $v_{max}$:

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

Using the maximum kinetic energy in Joules:

$$v\_\{max\}\; =\; \backslash sqrt\{\backslash frac\{2\; \backslash times\; (0.5516\; \backslash times\; 10^\{-19\}\; \backslash text\{\; J\})\}\{9.1\; \backslash times\; 10^\{-31\}\; \backslash text\{\; kg\}\}\}\; =\; \backslash sqrt\{\backslash frac\{1.1032\; \backslash times\; 10^\{-19\}\}\{9.1\; \backslash times\; 10^\{-31\}\}\}\; \backslash text\{\; m/s\}$$

$$v\_\{max\}\; =\; \backslash sqrt\{1.2123\; \backslash times\; 10^\{11\}\}\; \backslash text\{\; m/s\}\; \backslash approx\; 3.48\; \backslash times\; 10^5\; \backslash text\{\; m/s\}$$

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