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

Solution:

(a) Given the potential difference across the evacuated tube, $V\; =\; 500\; \backslash text\{\; V\}$, and the specific charge of an electron, $e/m\; =\; 1.76\; \backslash times\; 10^\{11\}\; \backslash text\{\; C\; kg\}^\{-1\}$. We assume the electrons are emitted with negligible initial velocity. The kinetic energy gained by an electron when accelerated through a potential difference V is equal to the work done by the electric field on the electron, which is $eV$. Using the classical kinetic energy formula $KE\; =\; \backslash frac\{1\}\{2\}mv^2$, we can set up the equation:

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

To find the speed $v$, we rearrange the formula:

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

Now, substitute the given values:

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

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

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

Therefore, the speed of each emitted electron is approximately $1.327\; \backslash times\; 10^7$ m/s.

(b) For a collector potential of $V\; =\; 10\; \backslash text\{\; MV\}\; =\; 10\; \backslash times\; 10^6\; \backslash text\{\; V\}$, we use the same formula derived in part (a):

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

Substituting the new potential value:

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

$v\; =\; \backslash left(20\; \backslash times\; 10^6\; \backslash times\; 1.76\; \backslash times\; 10^\{11\}\backslash right)^\{\backslash frac\{1\}\{2\}\}\; =\; \backslash left(35.2\; \backslash times\; 10^\{17\}\backslash right)^\{\backslash frac\{1\}\{2\}\}\; =\; \backslash left(3.52\; \backslash times\; 10^\{18\}\backslash right)^\{\backslash frac\{1\}\{2\}\}$

$v\; \backslash approx\; 1.88\; \backslash times\; 10^9\; \backslash text\{\; m/s\}$

This 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), which is physically impossible. The issue arises because the classical kinetic energy formula $KE\; =\; \backslash frac\{1\}\{2\}mv^2$ is only valid 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 the mass of the electron effectively increases. Therefore, the formula needs to be modified using relativistic mechanics. In the relativistic limit, the total energy $E$ is given by $E\; =\; mc^2$, where $m$ is the relativistic mass, and the kinetic energy is $K\; =\; mc^2\; -\; m\_0c^2$, where $m\_0$ is the rest mass. The relationship between relativistic mass $m$ and rest mass $m\_0$ is $m\; =\; \backslash frac\{m\_0\}\{\backslash left(1\; -\; \backslash frac\{v^2\}\{c^2\}\backslash right)^\{\backslash frac\{1\}\{2\}\}\}$.

Solution:

(a) We are given the speed of the electrons in the beam, $v\; =\; 5.20\; \backslash times\; 10^6$ m/s, the magnetic field strength $B\; =\; 1.30\; \backslash times\; 10^\{-4\}$ T, and the specific charge of the electron $e/m\; =\; 1.76\; \backslash times\; 10^\{11\}$ C kg^-1. Since the magnetic field is normal to the beam velocity, the magnetic force provides the centripetal force required for circular motion.

The magnetic force on an electron is given by $F\_B\; =\; evB$ (since $\backslash sin(90^\backslash circ)\; =\; 1$).

The centripetal force required for circular motion is $F\_c\; =\; \backslash frac\{mv^2\}\{r\}$, where $r$ is the radius of the circular path.

Equating the magnetic force to the centripetal force:

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

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

$$r\; =\; \backslash frac\{mv\}\{eB\}\; =\; \backslash frac\{v\}\{(e/m)B\}$$

Now, substitute the given values:

$r\; =\; \backslash frac\{5.20\; \backslash times\; 10^6\}\{(1.76\; \backslash times\; 10^\{11\})\; \backslash times\; (1.30\; \backslash times\; 10^\{-4\})\}$

$r\; =\; \backslash frac\{5.20\; \backslash times\; 10^6\}\{2.288\; \backslash times\; 10^7\}$

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

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

(b) To determine if the formula used in part (a) 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. For a 20 MeV electron beam, the total energy $E$ is the sum of its rest mass energy and kinetic energy: $E\; =\; m\_0c^2\; +\; K\; =\; 0.511\; \backslash text\{\; MeV\}\; +\; 20\; \backslash text\{\; MeV\}\; =\; 20.511\; \backslash text\{\; MeV\}$.

Using the relativistic energy-momentum relation $E^2\; =\; (pc)^2\; +\; (m\_0c^2)^2$, we can find the momentum $p$. Since 20 MeV is significantly larger than the rest mass energy (0.511 MeV), the electron is highly relativistic ($K\; \backslash gg\; m\_0c^2$). This means its speed will be very close to the speed of light ($c$).

The classical formula $r\; =\; mv/eB$ assumes that the mass $m$ is constant. However, in relativistic mechanics, the relativistic mass $m$ increases with speed according to $m\; =\; \backslash frac\{m\_0\}\{\backslash left(1\; -\; \backslash frac\{v^2\}\{c^2\}\backslash right)^\{\backslash frac\{1\}\{2\}\}\}$. At speeds close to $c$, this increase in mass is substantial, and the classical formula is no longer accurate.

Therefore, the formula used in part (a) is NOT valid for calculating the radius of the path of a 20 MeV electron beam. The formula needs to be modified by using the relativistic momentum $p$ instead of $mv$. The relativistic relation for the radius becomes $r\; =\; \backslash frac\{p\}\{eB\}$, where $p$ is the relativistic momentum, which can be calculated from the relativistic energy and rest mass energy.