At `10^(@)C` tempreture ,de-Broglie wave length of atom is 0.4 Å.If temperatire of atom is increased by `30^(@)C` ,What will be change in de-Broglie of atom?

Options

A. decreases `10^(2)Å`
B. decreases `2xx10^(-1)Å`
C. increase `10^(-2)` Å
D. increase `2xx10^(-2)Å`

Answer

B

Explanation

`T_(1)=10+273=283K`
`T_(2)=40+273=313K`
`lambda_(1)=0.4 Å`
`lambda=(h)/(sqrt(2mE))`
but,`E=(3)/(2)KT` so `lambda =(h)/(sqrt(3mKT))`
`therefore lambda prop (1)/(sqrt(T))`
`therefore (lambda_(2))/(lambda_(1))=sqrt(T_(1)/(T_(2)))`
`therefore (lambda_(2))/(0.4xx10^(-10))=sqrt((283)/(313))`
`therefore (lambda_(2))/(0.4xx10^(-10))=0.951`
`therefore lambda_(2)=0.951xx0.4xx10^(-10)`
`therefore lambda_(2)0.38 Å`
`therefore "wavelength" =lambda_(2)-lambda_(1)`