An electron is accelerated through a potential difference of 45.5 volt. The velocity acquired by it is (in $m s^{- 1}$ ) :

10^{6}

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

zero

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

4 \times 10^{6}

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

4 \times 10^{4}

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

Open in App

Solution

The correct option is C $4 \times 10^{6}$
$\frac{1}{2} m v^{2} = E$
$⟹ v = \sqrt{\frac{2 E}{m}}$
$= \sqrt{\frac{2 \times 45.5 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}}$
$= 4 \times 10^{6} m / s$

Suggest Corrections

Similar questions

Q. An electron is accelerated through potential difference of 45.5 volt. The velocity acquired by it is (in

{m s}^{- 1}

) :

An electron in accelerated through a potential difference of 45.5 V. The velocity acquired by it is $(i n m s^{- 1})$

Q. An electron is accelerated through a potential difference of

500 v o l t

. The velocity acquired by the electron is:

Q. An electron is accelerated through a potential difference of 1000 volt. The velocity acquired by the electron is

An electron is accelerated through a potential difference of $200$ volts. If $\frac{e}{m}$ ratio for the electron be $1.6 \times 10^{11}$ coulomb/kg, the velocity acquired by the electron will be :

Join BYJU'S Learning Program

An electron is accelerated through a potential difference of 45.5 volt. The velocity acquired by it is (in ms−1) :

The correct option is C 4×10612mv2=E⟹v=√2Em=√2×45.5×1.6×10−199.1×10−31=4×106m/s

An electron is accelerated through a potential difference of 45.5 volt. The velocity acquired by it is (in $m s^{- 1}$ ) :

The correct option is C $4 \times 10^{6}$
$\frac{1}{2} m v^{2} = E$
$⟹ v = \sqrt{\frac{2 E}{m}}$
$= \sqrt{\frac{2 \times 45.5 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}}$
$= 4 \times 10^{6} m / s$