An electron is accelerated under a potential difference of 182 V.Find the maximum velocity of electrons that will be existing between two parallel plates? The field between the plates, taking $(g = 10 m s^{- 2})$ is

8 \times 10^{6} m / s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

16 \times 10^{6} m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 \times 10^{6} m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.68 \times 10^{6} m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $8 \times 10^{6} m / s$
$\frac{1}{2} m v^{2} = e V \Rightarrow v = \sqrt{\frac{2 e V}{m}} = \sqrt{\frac{1.6 \times 10^{- 19} \times 2 \times 182}{9.1 \times 10^{- 21}}} \Rightarrow v = 8 \times 10^{6} m / s$

Suggest Corrections

Similar questions

Q. Light of frequency

ν = 5 ν_{0}

falls on a certain metal, then maximum velocity of electrons emitted is

8 \times 10^{6}

m/s where

ν_{0}

is threshold frequency of metal. If

ν = 2 ν_{0}

, then the maximum velocity of photoelectrons will be

Q. The speed of an electron having a wavelength of

10^{- 10} m

The circumference of the first Bohr orbit in H atom is $3.322 \times 10^{- 10} m$ . What is the velocity of the electron of this orbit?

The threshold wavelength for certain metal is . When light of wavelength $\frac{λ_{0}}{2}$ is incident on it, the maximum velocity of photo electrons is $10^{6} m / s$ . If the wavelength of incident light is changed to $\frac{λ_{0}}{5}$ the maximum velocity of photo electrons is

What is the speed of electron in the second Bohr's orbit of Hydrogen atom? $(T a k e v_{0} = 2.2 \times 10^{6} m / s)$ .

Join BYJU'S Learning Program

An electron is accelerated under a potential difference of 182 V.Find the maximum velocity of electrons that will be existing between two parallel plates? The field between the plates, taking (g=10ms−2) is

The correct option is A 8×106m/s12mv2=eV⇒v=√2eVm=√1.6×10−19×2×1829.1×10−21⇒v=8×106m/s

An electron is accelerated under a potential difference of 182 V.Find the maximum velocity of electrons that will be existing between two parallel plates? The field between the plates, taking $(g = 10 m s^{- 2})$ is

The correct option is A $8 \times 10^{6} m / s$
$\frac{1}{2} m v^{2} = e V \Rightarrow v = \sqrt{\frac{2 e V}{m}} = \sqrt{\frac{1.6 \times 10^{- 19} \times 2 \times 182}{9.1 \times 10^{- 21}}} \Rightarrow v = 8 \times 10^{6} m / s$