Question

A hydrogen molecule and helium atom are moving with the same velocity. Then the ratio of their de-broglie wavelength is

Answer 1

Answer:

Lambda( H2)= 2 lambda(He)

Explanation:

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De Broglie wavelength given by = h/MV

since V is same , h constant also same

we have lambda(¥) = 1/M

$\frac{ \gamma 1}{ \gamma 2} = \frac{m2}{m1}$

mass of helium 4 , mass of hydrogen 2

$\frac{ \gamma 1}{ \gamma 2} = \frac{4}{2 } = 2$

$\gamma 1 = 2 \gamma 2$

Answer 2

Answer:

the correct answer is 1/2