A beam of electrons of velocity 3 × 107 ms -1 is deflected 1.5 mm is passing 10 cm through an electric field of 1800 Vm -1 perpendicular to their path. The value of (e/m) for electron is

Question

A beam of electrons of velocity 3 × 107 ms -1 is deflected 1.5 mm is passing 10 cm through an electric field of 1800 Vm -1 perpendicular to their path. The value of (e/m) for electron is (A) 1.78 × 1011 C kg -1 (B) 2 × 1011 Ckg -1 (C) 1.5 × 1011 C kg -1 (D) 3.5 × 1011 C kg -1

Tardigrade Admin · Accepted Answer

As, y=(1/2) a t2 =(1/2) (e E/m) (x2/v2) or (e/m) =(2 y v2/E x2) =(2 × 1.5 × 10-3 ×(3 × 107)2/1800 ×(0.1)2) =1.5 × 1011 Ckg -1

Q. A beam of electrons of velocity $3 \times 1 0^{7} m s^{- 1}$ is deflected $1.5 mm$ is passing $10 c m$ through an electric field of $1800 V m^{- 1}$ perpendicular to their path. The value of $\frac{e}{m}$ for electron is

$1.78 \times 1 0^{11} C k g^{- 1}$

$2 \times 1 0^{11} C k g^{- 1}$

$1.5 \times 1 0^{11} C k g^{- 1}$

$3.5 \times 1 0^{11} C k g^{- 1}$

As, $y = \frac{1}{2} a t^{2} = \frac{1}{2} \frac{e E}{m} \frac{x ^{2}}{v ^{2}}$
or $\frac{e}{m} = \frac{2 y v ^{2}}{E x ^{2}}$
$= \frac{2 \times 1.5 \times 1 0 ^{- 3} \times ( 3 \times 1 0 ^{7} ) ^{2}}{1800 \times ( 0.1 ) ^{2}}$
$= 1.5 \times 1 0^{11} C k g^{- 1}$

Q. A beam of electrons of velocity 3×107ms−1 is deflected 1.5mm is passing 10cm through an electric field of 1800Vm−1 perpendicular to their path. The value of me​ for electron is

1.78×1011Ckg−1

2×1011Ckg−1

1.5×1011Ckg−1

3.5×1011Ckg−1