Question

A helium molecule is moving with a velocity of 2.40 *10^2 m/s at 300K. The de-broglie wave length is about:

Answer 1

☆☆ AS-SALAMU-ALYKUM ☆☆

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GIVEN :-

VELOCITY = 2.40 × 10^2 m/s

TEMPERATURE = 300K

FORMULA

lamda = h ÷ mv

h is plancks constant =6.62×10^_34

m is mass = molar mass of helium molecule is 4g

SOLUTION :-

lamda = h÷mv

lamda = 6.62×10^_34 ÷ 4 × 2.40 ×10^2

LAMDA = 3.97 ×10^_34 +2

lamda = 3.97 × 10^_32

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