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Question: Relativistic corrections become necessary when the expression for the kinetic energy\(\frac{1}{2}my^...

Relativistic corrections become necessary when the expression for the kinetic energy12my2\frac{1}{2}my^{2}, becomes comparable with mc2 , where m is the mass of the particle . At what de Broglie wavelength will relativistic corrections becomes important for an electron?

A

λ=1nm\lambda = 1nm

B

λ=10nm\lambda = 10nm

C

λ=10−1nm\lambda = 10^{- 1}nm

D

λ=10−4nm\lambda = 10^{- 4}nm

Answer

λ=10−4nm\lambda = 10^{- 4}nm

Explanation

Solution

: Here h=6.63×10−34jsm=9×10−31kgh = 6.63 \times 10^{- 34}jsm = 9 \times 10^{- 31}kg

Velocity of electron,

v=hmλv = \frac{h}{m\lambda}…… (i)

(i) When λ1=1nm=10−9m\lambda_{1} = 1nm = 10^{- 9}m

v1=6.63×10−349×10−31×10−9=0.74×106ms−1=106ms−1v_{1} = \frac{6.63 \times 10^{- 34}}{9 \times 10^{- 31} \times 10^{- 9}} = 0.74 \times 10^{6}ms^{- 1} = 10^{6}ms^{- 1}

(ii) Whenλ2=10nm=10×10−9m=10−8m\lambda_{2} = 10nm = 10 \times 10^{- 9}m = 10^{- 8}m

v2=6.63×10−349×10−31×10−8=105ms−1v_{2} = \frac{6.63 \times 10^{- 34}}{9 \times 10^{- 31} \times 10^{- 8}} = 10^{5}ms^{- 1}

(iii) Whenλ3=10−1nm=10−1×10−9m=10−10m\lambda_{3} = 10^{- 1}nm = 10^{- 1} \times 10^{- 9}m = 10^{- 10}m

v3=6.63×10−349×10−31×10−10=107ms−1v_{3} = \frac{6.63 \times 10^{- 34}}{9 \times 10^{- 31} \times 10^{- 10}} = 10^{7}ms^{- 1}

(iv) Whenλ4=10−4nm=10−4×10−9m=10−13m\lambda_{4} = 10^{- 4}nm = 10^{- 4} \times 10^{- 9}m = 10^{- 13}m

v4=6.63×10−349×10−31×10−13=1010ms−1v_{4} = \frac{6.63 \times 10^{- 34}}{9 \times 10^{- 31} \times 10^{- 13}} = 10^{10}ms^{- 1}

Asv4v_{4}is greater than the velocity of light (=3×108ms−1)( = 3 \times 10^{8}ms^{- 1}) hence relativistic correction Is needed for λ=10−4nm\lambda = 10^{- 4}nm