Question

The de broglie wavelength of an electron travelling with 10 of velocity of light is equal to

Answer 1

Answer:

$lamda = h \div p$

here h =3.34 ×10^-34

p=m.v

m= 9.1×10^-31

v=3×10^8

hope it helps to calculate further

Answer 2

de broglie wavelength has a formula

$h \div mv$

where h is planck' s constant and

m is mass of e-(9.1 ×10–³¹ kg)

v is velocity which is given

10 times the velocity of light that is

10 times the velocity of light that is 3×10^8×10 that is 3× 10^9 m /sec

hence having the value of

planck's constant as 6.63× 10–³⁴ js

wavelength = 6.63× 10–³⁴/ (9.1 ×10–³¹×3× 10^9)

wavelength = 6.63 × 10–¹² /3× 9.1

0.242 × 10–¹² m Or 2.42× 10-¹³ m

this is ur ans

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