Test Index

NCERT Class XII Chapter
Dual Nature of Radiation and Matter
Questions With Solutions

© examsnet.com
Question : 20 of 37
Marks: +1, -0
(a) Estimate the speed with which electrons emitted from a heated cathode of an evacuated tube impinge on the anode maintained at a potential difference of 500 V with respect to the cathode. 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 × 1011 C kg−110^{11}\,\mathrm{C}\,\mathrm{kg}^{-1}.
(b) Use the same formula you employ in (a) to obtain electron speed for an anode potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?
Solution:  
(a) Energy of accelerated electron E = 500 eV
Specific charge, e/m = 1.76 × 1011 C kg–1,
Kinetic energy E = 12mv2\frac{1}{2} mv^{2} = eV
∴ v = 2V(em)\sqrt{2V\left(\frac{e}{m}\right)} =
2×1.76×1011×500\sqrt{2 \times 1.76 \times 10^{11} \times 500}
= 1.33 × 10710^{7} m −1^{-1}
(b) For anode potential V = 10 MV = 10710^{7} V
Speed v =
2×1.76×1011×107\sqrt{2 \times 1.76 \times 10^{11} \times 10^{7}}
= 1.88 × 10910^{9} m s−1\mathrm{s}^{-1}
This speed of electron is impossible. Since nothing can move with a speed greater than speed of light (c = 3 × 108 m s−110^{8}\,\mathrm{m}\,\mathrm{s}^{-1}).
The formula for kinetic energy E = 1/2 mv2mv^{2} is valid only for v << c. For the situation when speed v is comparable to speed of light c, we use relativistic formula.
The relativistic mass is given by
m = m01−v2c2\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} , where m0m_{0} is rest mass
Also total energy is taken as
E2E^{2} = c2p2+m02c4c^{2}p^{2}+m_{0}^{2}c^{4} or E2E^{2} = c2[p2+m02c2]c^{2}[p^{2}+m_{0}^{2}c^{2}]
E2E^{2} = c2[m02v21−v2c2+m02c2]c^{2}\left[\frac{m_{0}^{2}v^{2}}{1-\frac{v^{2}}{c^{2}}}+m_{0}^{2}c^{2}\right] ⇒ E = m0c21−v2c2\frac{m_{0}c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}
Now, K = E – m0c2m_{0}c^{2}
∴ eV = m0c2[11−v2c2−1]m_{0}c^{2}\left[\frac{1}{\sqrt{\frac{1-v^{2}}{c^{2}}}}-1\right]
or eVm0c2\frac{eV}{m_{0}c^{2}} = 11−v2c2\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} - 1
Substituting the values
1.6×10−19×10×1069.1×10−31×(3×108)2\frac{1.6 \times 10^{-19} \times 10 \times 10^{6}}{9.1 \times 10^{-31} \times (3 \times 10^{8})^{2}} + 1 = 11−v2c2\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}
19.536 + 1 = 11+vc2\frac{1}{\sqrt{1+\frac{v}{c^{2}}}} or 1 - v2c2\frac{v^{2}}{c^{2}} = 0.00237
v2c2\frac{v^{2}}{c^{2}} = 0.997 or speed v = 0.997\sqrt{0.997} c = 0.999 c
© examsnet.com
Go to Question:
Prev Question
Next Question