JEE Main 26 Feb 2021 Shift 1 Solved Paper

Given, work done by gas molecule,
W=α2βe−βx2∕kT
Here, x is displacement, k is Boltzmann constant, α and β are constants and T is temperature.
Dimensional formula of [W]=[ML2T−2]
∴ Dimensions of [α2β]=[ML2T−2]
⇒‌‌α=[‌

ML2T−2

β

]1∕2 . . . (i)
The term [e−βx2∕kT] should be dimensionless, i.e. [M0L0T0].
⇒‌‌[‌

βx2

kT

]=[M0L0T0]
⇒‌‌[β]=‌

[k][T]

[x2]

. . . (ii)
‌ Energy of gaseous molecule ‌(E)=‌

7

2

kT
[k]=[E]∕[T]=[ML2T−2K−1]
Substituting the value of k in Eq. (ii), we get
[β]=‌

[ML2T−2K−1][K]

[L2]

=[MT−2]
Substituting the value of β in Eq. (i), we get
[α]={‌

[ML2T−2]

[MT−2]

}1∕2=[ML1T0]