Question

the energy of helium atom at 300k is

38.8 meV
3.88 MeV
0.388 eV
3.88 eV​

Answer 1

Answer:

3.88MeV is the correct answer

Answer 2

The energy of helium atom at 300k is 0.388 eV.

Here, according to the given information, we are given that,

The temperature is equal to 300K.

Now, in order to find the average thermal energy of the Helium atom, we need to utilize the formula of average thermal energy that is,

$\frac{3}{2} kT$, where k is called the Boltzmann constant and T is the temperature at which the average thermal energy of the helium atom needs to be calculated and that temperature is given here as 300K.

Now, we know that the Boltzmann constant that is k, is given as,

k = 1.38 × $10^{-23}$ m²kg$s^{-2}$$K^{-1}$

Now, putting the values in the above formula we get,

$\frac{3}{2} kT\\\\=\frac{3.(1.38).10^{-23}.300}{2.(1.602).(10^{-19)} }$

= 0.388 eV

Hence, the energy of helium atom at 300k is 0.388 eV.

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