Question

When a cathode ray tube is operated at $2912$ volt, the velocity of electrons is $3.2\times {10}^7$m/sec. Find the velocity of cathhode ray if the tube is operated at $5824$ volt:

• A. $2.4\times {10}^{7}m/s$
• B. $5.2\times {10}^{7}m/s$
• C. $4.525\times {10}^{7}m/s$
• D. $6.2\times {10}^{6}m/s$

Answer 1

The correct option is C $4.525\times {10}^{7}m/s$

Given,

$V_1=2912V$, $v_1=3.2\times {10}^{7}m/s$

$V_2=5824V$, $v_2=?$

The wavelength and voltage relation is given by,

$\lambda =\frac{12.27}{\sqrt{V}}\times {10}^{-10}m$. . . . .(1)

The energy momentum relation is

$\lambda =\frac{h}{mv}$. . . . . . .(2)

From equation (1) and (2), we can say that

$\frac{v_1}{v_2}=\sqrt{\frac{V_1}{V_2}}$

$v_2=v_1\sqrt{\frac{V_2}{V_1}}=2.3\times {10}^7\sqrt{\frac{5824}{2912}}$

$V_2=4.52\times {10}^{7}m/s$

The correct option is C.