Find the velocity of an electron having an energy of 1280 electron volt. $m = 9.1 \times 10^{- 31} k g$ .

Open in App

Solution

$E = \frac{1}{2} m v^{2}$

$⟹ v = \sqrt{\frac{2 E}{m}}$

$⟹ v = \sqrt{\frac{2 \times 1280 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}}$

$⟹ v = 2.12 \times 10^{7} m / s$

Suggest Corrections

Similar questions

Q. An electron is accelerated from rest through a P.D. of 1600 volt. Find the velocity acquired by the electron. Mass of electron =

9.1 \times 10^{- 31} k g .

The kinetic energy of an electron is $4.55 \times 10^{- 25} J$ . The mass ofelectron is $9.1 \times 10^{- 31} k g$ . Calculate velocity of the electron.

Q. The kinetic energy of an electron is

4.55 \times 10^{- 25} J

.The mass of electron is

9.1 x 10^{- 31} kg

.The velocity of the electron is .

Q. Calculate the wavelength of an electron if its kinetic energy is

3.0 \times 10^{- 25} J

. Mass of an electron is

9.1 \times 10^{- 31} K g

Q. The de Broglie wavelength of an electron having

320 eV

of energy is nearly

(1 eV = 1.6 \times 10^{- 19} J

, mass of electron

= 9.1 \times 10^{- 31} kg

, Planck's constant

= 6.6 \times 10^{- 34} Js)

Join BYJU'S Learning Program

Find the velocity of an electron having an energy of 1280 electron volt. m=9.1×10−31kg.

E=12mv2⟹v=√2Em⟹v=√2×1280×1.6×10−199.1×10−31⟹v=2.12×107m/s

Find the velocity of an electron having an energy of 1280 electron volt. $m = 9.1 \times 10^{- 31} k g$ .

$E = \frac{1}{2} m v^{2}$

$⟹ v = \sqrt{\frac{2 E}{m}}$

$⟹ v = \sqrt{\frac{2 \times 1280 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}}$

$⟹ v = 2.12 \times 10^{7} m / s$