Solveeit Logo
Login

Question

Question: An electron is moving with a kinetic energy of 4.55 × 10\(- 25\) J. What will be de-Broglie waveleng...

An electron is moving with a kinetic energy of 4.55 × 10−25- 25 J. What will be de-Broglie wavelength for this electron

A

5.28 × 10−7m1{0 - 7}_{}m

B

7.28 ×10−7m10_{}^{- 7}m

C

2 × 10−10m_{}^{- 10}m

D

3 × 10−5m10_{}^{- 5}m

Answer

7.28 ×10−7m10_{}^{- 7}m

Explanation

Solution

KE=12mv2=4.55×10−25= \frac { 1 } { 2 } m v ^ { 2 } = 4.55 \times 10 ^ { - 25 } J

v2=2×4.55×10−259.1×10−31=1×106v ^ { 2 } = \frac { 2 \times 4.55 \times 10 ^ { - 25 } } { 9.1 \times 10 ^ { - 31 } } = 1 \times 10 ^ { 6 }; v=103m/sv = 103_{}m/s

De-Broglie wavelength

λ=hmv=6.626×10−349.1×10−31×103=7.28×10−7 m\lambda = \frac { h } { m v } = \frac { 6.626 \times 10 ^ { - 34 } } { 9.1 \times 10 ^ { - 31 } \times 10 ^ { 3 } } = 7.28 \times 10 ^ { - 7 } \mathrm {~m}