An electron is accelerated through a potential difference of $10, 000 V$ . Its de Broglie wavelength is (nearly) $(m_{e} = 9 \times 10^{- 31} k g)$

12.2 \times 10^{- 13} m

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

12.2 \times 10^{- 12} m

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

12.2 \times 10^{- 14} m

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

12.2 n m

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

Open in App

Solution

The correct option is B $12.2 \times 10^{- 12} m$
$λ = \sqrt{\frac{150}{V}} A^{0} = \sqrt{\frac{150}{10^{4}}} A^{0} = 12.2 \times 10^{- 12} m$

Suggest Corrections

Similar questions

10, 000 V

(m_{e} = 9 \times 10^{- 31} k g)

An electron has kinetic energy $2.8 \times 10^{- 23} J$ de-Broglie wavelength will be nearly $m_{e} = 9.1 \times 10^{- 31} k g$

An electron has kinetic energy $2.8 \times 10^{- 23} J$ . De-Broglie wavelength will be nearly

$(m e = 9.1 \times 10^{- 31} k g)$

m_{e} = 9.1 \times 10^{- 31} k g, e = 1.6 \times 10^{- 19} C,

h = 6.6 \times 10^{- 31} J s

Join BYJU'S Learning Program