CBSE Class 12 Physics NCERT Solutions: Chapter 20

Solution:

Part (a): Calculating electron speed at 500 V

Given:

Potential difference, $V\; =\; 500\; \backslash text\{\; V\}$

Specific charge of an electron, $\backslash frac\{e\}\{m\}\; =\; 1.76\; \backslash times\; 10^\{11\}\; \backslash text\{\; C\; kg\}^\{-1\}$

When electrons are accelerated by a potential difference $V$, the potential energy lost is converted into kinetic energy. Assuming the initial velocity of the electrons is negligible, we can write:

$$KE\; =\; \backslash frac\{1\}\{2\}mv^2\; =\; eV$$

Where $m$ is the mass of the electron, $v$ is its final speed, $e$ is the charge of the electron, and $V$ is the potential difference.

We can rearrange this formula to solve for the speed $v$:

$$v\; =\; \backslash left(\backslash frac\{2eV\}\{m\}\backslash right)^\{\backslash frac\{1\}\{2\}\}\; =\; \backslash left(2V\; \backslash times\; \backslash frac\{e\}\{m\}\backslash right)^\{\backslash frac\{1\}\{2\}\}$$

Now, substitute the given values:

$v\; =\; \backslash left(2\; \backslash times\; 500\; \backslash text\{\; V\}\; \backslash times\; 1.76\; \backslash times\; 10^\{11\}\; \backslash text\{\; C\; kg\}^\{-1\}\backslash right)^\{\backslash frac\{1\}\{2\}\}$

$v\; =\; \backslash left(1000\; \backslash times\; 1.76\; \backslash times\; 10^\{11\}\backslash right)^\{\backslash frac\{1\}\{2\}\}\; =\; \backslash left(1.76\; \backslash times\; 10^\{14\}\backslash right)^\{\backslash frac\{1\}\{2\}\}\; \backslash text\{\; m/s\}$

$v\; \backslash approx\; 1.327\; \backslash times\; 10^7\; \backslash text\{\; m/s\}$

Thus, the speed of the electrons impinging on the collector is approximately $1.327\; \backslash times\; 10^7$ m/s.

Part (b): Electron speed at 10 MV and modification of the formula

Given:

Collector potential, $V\; =\; 10\; \backslash text\{\; MV\}\; =\; 10\; \backslash times\; 10^6\; \backslash text\{\; V\}$

Using the same non-relativistic formula as in part (a):

$v\; =\; \backslash left(2V\; \backslash times\; \backslash frac\{e\}\{m\}\backslash right)^\{\backslash frac\{1\}\{2\}\}$

$v\; =\; \backslash left(2\; \backslash times\; 10\; \backslash times\; 10^6\; \backslash text\{\; V\}\; \backslash times\; 1.76\; \backslash times\; 10^\{11\}\; \backslash text\{\; C\; kg\}^\{-1\}\backslash right)^\{\backslash frac\{1\}\{2\}\}$

$v\; =\; \backslash left(20\; \backslash times\; 10^6\; \backslash times\; 1.76\; \backslash times\; 10^\{11\}\backslash right)^\{\backslash frac\{1\}\{2\}\}\; =\; \backslash left(35.2\; \backslash times\; 10^\{17\}\backslash right)^\{\backslash frac\{1\}\{2\}\}\; \backslash text\{\; m/s\}$

$v\; =\; \backslash left(3.52\; \backslash times\; 10^\{18\}\backslash right)^\{\backslash frac\{1\}\{2\}\}\; \backslash text\{\; m/s\}\; \backslash approx\; 1.88\; \backslash times\; 10^9\; \backslash text\{\; m/s\}$

Analysis of the result:

The calculated speed $1.88\; \backslash times\; 10^9$ m/s is greater than the speed of light ($c\; \backslash approx\; 3\; \backslash times\; 10^8$ m/s). This is physically impossible, indicating that the formula used is not valid under these conditions.

Modification of the formula:

The formula $KE\; =\; \backslash frac\{1\}\{2\}mv^2$ is based on classical mechanics and is only accurate for speeds much smaller than the speed of light ($v\; \backslash ll\; c$). When speeds approach the speed of light, relativistic effects become significant, and we must use Einstein's theory of special relativity.

In relativistic mechanics, the total energy $E$ of a particle is given by $E\; =\; mc^2$, where $m$ is the relativistic mass, and the kinetic energy $K$ is the difference between the total energy and the rest energy ($m\_0c^2$):

$K\; =\; E\; -\; m\_0c^2\; =\; mc^2\; -\; m\_0c^2$

The relativistic mass $m$ is related to the rest mass $m\_0$ by:

$m\; =\; \backslash frac\{m\_0\}\{\backslash sqrt\{1\; -\; \backslash frac\{v^2\}\{c^2\}\}\}$

The kinetic energy gained by the electron from the potential difference $V$ is $eV$. Therefore, the relativistic equation to solve for $v$ would be:

$$eV\; =\; \backslash left(\backslash frac\{m\_0\}\{\backslash sqrt\{1\; -\; \backslash frac\{v^2\}\{c^2\}\}\}\backslash right)c^2\; -\; m\_0c^2$$

This equation needs to be solved for $v$ when $eV$ is large enough to cause relativistic speeds.

Solution:

Part (a): Radius of the circular path at $5.20\; \backslash times\; 10^6$ m/s

Given:

Electron speed, $v\; =\; 5.20\; \backslash times\; 10^6$ m/s

Magnetic field strength, $B\; =\; 1.30\; \backslash times\; 10^\{-4\}$ T

Specific charge of electron, $\backslash frac\{e\}\{m\}\; =\; 1.76\; \backslash times\; 10^\{11\}$ C kg^-1

When a charged particle moves in a magnetic field perpendicular to its velocity, it experiences a Lorentz force given by $F\; =\; evB$. This force acts as the centripetal force, causing the particle to move in a circular path. The centripetal force is given by $F\_c\; =\; \backslash frac\{mv^2\}\{R\}$, where $R$ is the radius of the circular path.

Equating the magnetic force and the centripetal force:

$$evB\; =\; \backslash frac\{mv^2\}\{R\}$$

We can rearrange this equation to solve for the radius $R$:

$$R\; =\; \backslash frac\{mv\}\{eB\}\; =\; \backslash frac\{v\}\{\backslash left(\backslash frac\{e\}\{m\}\backslash right)B\}$$

Now, substitute the given values:

$R\; =\; \backslash frac\{5.20\; \backslash times\; 10^6\; \backslash text\{\; m/s\}\}\{\backslash left(1.76\; \backslash times\; 10^\{11\}\; \backslash text\{\; C\; kg\}^\{-1\}\backslash right)\; \backslash times\; \backslash left(1.30\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\}\backslash right)\}$

$R\; =\; \backslash frac\{5.20\; \backslash times\; 10^6\}\{1.76\; \backslash times\; 1.30\; \backslash times\; 10^\{7\}\}\; \backslash text\{\; m\}$

$R\; =\; \backslash frac\{5.20\; \backslash times\; 10^6\}\{2.288\; \backslash times\; 10^\{7\}\}\; \backslash text\{\; m\}$

$R\; \backslash approx\; 0.227\; \backslash text\{\; m\}$

The radius of the circle traced by the electron beam is approximately 0.227 meters.

Part (b): Validity of the formula for a 20 MeV electron beam

To determine if the formula is valid for a 20 MeV electron beam, we first need to check the speed of such an electron. The rest mass energy of an electron ($m\_0$) is approximately 0.511 MeV.

The kinetic energy (KE) of the electron is given as 20 MeV.

The total relativistic energy $E$ is the sum of rest energy and kinetic energy:

$E\; =\; m\_0c^2\; +\; KE\; =\; 0.511\; \backslash text\{\; MeV\}\; +\; 20\; \backslash text\{\; MeV\}\; =\; 20.511\; \backslash text\{\; MeV\}$

The relativistic mass $m$ is related to total energy by $E\; =\; mc^2$. The relativistic speed $v$ can be found using the relation $E\; =\; \backslash frac\{m\_0c^2\}\{\backslash sqrt\{1\; -\; v^2/c^2\}\}$.

From $E\; =\; \backslash frac\{m\_0c^2\}\{\backslash sqrt\{1\; -\; v^2/c^2\}\}$, we get $\backslash sqrt\{1\; -\; v^2/c^2\}\; =\; \backslash frac\{m\_0c^2\}\{E\}$.

$\backslash sqrt\{1\; -\; v^2/c^2\}\; =\; \backslash frac\{0.511\; \backslash text\{\; MeV\}\}\{20.511\; \backslash text\{\; MeV\}\}\; \backslash approx\; 0.0249$

Squaring both sides: $1\; -\; \backslash frac\{v^2\}\{c^2\}\; =\; (0.0249)^2\; \backslash approx\; 0.00062$

$\backslash frac\{v^2\}\{c^2\}\; =\; 1\; -\; 0.00062\; =\; 0.99938$

$v^2\; =\; 0.99938\; c^2$

$v\; =\; \backslash sqrt\{0.99938\}\; c\; \backslash approx\; 0.99969\; c$

The speed of a 20 MeV electron is approximately 0.99969 times the speed of light. This is a highly relativistic speed ($v\; \backslash approx\; c$).

Conclusion on formula validity:

The formula used in part (a), $R\; =\; \backslash frac\{mv\}\{eB\}$, assumes that the mass $m$ is the rest mass $m\_0$ and that the classical kinetic energy formula applies. Since the speed is very close to the speed of light, this formula is **not valid**.

Modification required:

For relativistic speeds, the mass $m$ in the formula $R\; =\; \backslash frac\{mv\}\{eB\}$ must be replaced by the relativistic mass $m\_\{rel\}\; =\; \backslash frac\{m\_0\}\{\backslash sqrt\{1\; -\; v^2/c^2\}\}$. The formula becomes:

$$R\; =\; \backslash frac\{m\_\{rel\}v\}\{eB\}\; =\; \backslash frac\{m\_0v\}\{eB\backslash sqrt\{1\; -\; v^2/c^2\}\}$$

Alternatively, since the kinetic energy $K\; =\; eV$ (where V is the accelerating potential) is related to relativistic mass and momentum, and the magnetic force provides centripetal force, the calculation needs to be done using relativistic energy-momentum relations.

For a 20 MeV electron, the speed is so close to $c$ that the relativistic mass is significantly larger than the rest mass, and the classical formula will yield an incorrect radius.