At `200^(@)C` hydrogen molecules have velocity `2.4 xx 10^(5) cm s^(-1)` The de Brogelie waqvelength in this case is approximately

Options

A. `1 Å`
B. `1000 Å`
C. `100 Å`
D. `10 Å`

Answer

A

Explanation

`lambda = (h)/(mv) = (6.6 xx 10^(-27) erg s)/((2)/(N_(A)) xx 2.4 xx 10^(5) cm s^(-1)`
`("Mass of l molecule of" `H_(2) = (2)/(6.02 xx 10^(23)))`
`= (6.6 xx 10^(-27) xx 6.02 xx 10^(23))/( 2 xx 2.4 xx 10^(5))`
`= 8.27 xx 10^(-9) cm`
`= 0.8 xx 10^(8) cm = 0.8 Å = 1A`