Question

The de broglie wavelength of an electron moving with a velocity 1.5

Answer 1

Answer:

The de Broglie wavelength for electron = \lambda = h/p = c/\nuλ=h/p=c/ν

Given \lambda_e = \lambda_pλ

e

=λ

p

KE of electron = 1/2 \times m \times v^2 = 1/2 \times p \times v_e1/2×m×v

2

=1/2×p×v

e

Substituting for p in the above equation, we get

K.E of electron : \dfrac{hv_e}{2\lambda_e}

2λ

e

hv

e

\therefore \lambda_e = \cfrac{h \times v_e}{2 \times KE_e}∴λ

e

=

2×KE

e

h×v

e

The K.E of photon is given by : \dfrac{hc}{\lambda_p}

λ

p

hc

\therefore \lambda_p = \cfrac{h \times c}{ KE_p}∴λ

p

=

KE

p

h×c

Given that the wavelengths are equal .

Taking ratio we get : \dfrac{v_e}{2 \times KE_e}

2×KE

e

v

e

= \dfrac{c}{KE_p}

KE

p

c

Then,

\dfrac{KE_p}{KE_e} = \dfrac {2 \times c}{v_e} =

KE

e

KE

p

=

v

e

2×c

= \dfrac{2 \times 3 \times 10 ^8}{1.5 \times 10^8} = 4

1.5×10

8

2×3×10

8

= 4