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

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Question : 20
Total: 37
(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 × 1011Ckg–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 =
1
2
mv2 = eV
∴ v = √2V(e∕m) =
√2×1.76×1011×500
= 1.33 × 107 m ‌−1
(b) For anode potential V = 10 MV = 107 V
Speed v =
√2×1.76×1011×107
= 1.88 × 109 m s−1
This speed of electron is impossible. Since nothing can move with a speed greater than speed of light (c = 3 × 108ms–1).
The formula for kinetic energy E = 1/2 mv2 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 =
m0
√1−
v2
c2
 , where m0 is rest mass
Also total energy is taken as
E2 = c2p2+m02c4 or E2 = c2[p2+m02c2]
E2 = c2[
m02v2
1−
v2
c2
+m02c2] ⇒ E =
m0c2
√1−
v2
c2
Now, K = E – m0c2
∴ eV = m0c2[
1
√
1−v2
c2
−1]
or
eV
m0c2
 =
1
√1−
v2
c2
 - 1
Substituting the values
1.6×10−19×10×106
9.1×10−31×(3×108)2
 + 1 =
1
√1−
v2
c2
19.536 + 1 =
1
√1+
v1
c2
 or 1 -
v2
c2
 = 0.00237
v2
c2
 = 0.997 or speed v = √0.997 c = 0.999 c
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