(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of $500 \mathrm{~V}$ with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be $1.76 \times 10^{11} \mathrm{C} \mathrm{kg}^{-1}$. (b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of $10 \mathrm{MV}$. Do you see what is wrong? In what way is the formula to be modified?

Home » (a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be . (b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of . Do you see what is wrong? In what way is the formula to be modified?

(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of $500 \mathrm{~V}$ with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be $1.76 \times 10^{11} \mathrm{C} \mathrm{kg}^{-1}$ .
(b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of $10 \mathrm{MV}$ . Do you see what is wrong? In what way is the formula to be modified?

CBSE | Class 12 | Dual Nature of Radiation and Matter | NCERT | Physics

(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of $500 \mathrm{~V}$ with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be $1.76 \times 10^{11} \mathrm{C} \mathrm{kg}^{-1}$ .
(b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of $10 \mathrm{MV}$ . Do you see what is wrong? In what way is the formula to be modified?

(a) Potential difference of the evacuated tube is given as $500 \mathrm{~V}$

Specific charge of the electron is given as $\mathrm{e} / \mathrm{m}=1.76 \times 10^{11} \mathrm{C} \mathrm{kg}^{-1}$

Expression for Kinetic energy is,

$(1 / 2) \mathrm{mv}^{2}=\mathrm{eV}$

Now, Speed of the emitted electron can be calculated as,

(b) Collector potential is given as $\mathrm{V}=10 \mathrm{MV}=10 \times 10^{6} \mathrm{~V}$ .

Now, Speed of electron can be calculated as

$\mathrm{v}=(2 \mathrm{Ve} / \mathrm{m})^{1 / 2}$ $=\left(2 \times 10^{7} \times 1.76 \times 10^{11}\right)^{1 / 2}$ $=1.88 \times 10^{9} \mathrm{~m} / \mathrm{s}$

This answer is not correct. Since the value is greater than the speed of light. The expression $(1 / 2) \mathrm{mv}^{2}$ for energy should be used in the non -relativistic limit.i.e., $v \ll<$ c.