Question

Determine the approximate value of quantum number (n) for a hydrogen atom located in a
cubic box of 3 cm3
at 50 degrees centigrade. Mass of hydrogen atom is 9×10-28 g.

Answer 1

Here, the temperature of hydrogen atom is

t = 50 ^oC = 323 K

The thermal energy of the hydrogen atom is

E = k_bT,

Where the Boltzmann constant is given by

k_B = 1.38*10^-23 J/K

So,

E = 1.38*10^-23 *323 = 4.4574*10^-21 J

Now, if the temperature was 0K, the atom would be in the ground state of the potential well, given by

En = nºh?/8m

For ground state,

n = 3.

since

ni =n+n+n

Due to the temperature, T, the atom will be in the excited state m.

Where

Em = m 2 /8mL2 = 3212 78ml + PT

So,

m2 = 32 + E + 8mL/h

Here,

L = 3cm = 3 * 10-6m

So,

L? = (3 * 10-652/3 m

The mass of hydrogen atom given here is wrong. The actual value is

m = 1.67*10^-28 kg

plank's constant, h = 6.626*10^-34 m² kg/s

So,

m2 = 32 + 4.4574 * 10-21 * 8 * 1.67 * 10-28 * (3 * 10-6)2/3 -= 2.82 * 1015 (6.626 * 10-34)2

m = 5.312 * 107

This value of m is very high. This is expected, because this temperature of 50 degree celsius is very high compared to the ground state energy of a potential well.

At this temperature, it is very hard to confine a hydrogen atom in a perfect potential well.