Q.Derive the relation between coefficient of absorption, coefficient of reflection and coefficient of transmission. Compare the r.m.s. speed of hydrogen molecule at 127°C with r.m.s. speed of oxygen molecule at 27°C, given that molecular masses of hydrogen and oxygen are 2 and 32 respectively.

Splitting incident radiant energy into absorbed, reflected, and transmitted parts and dividing by the total gives $a+r+t=1$; the r.m.s.-speed ratio follows directly from $v_{rms}\propto\sqrt{T/M}$.

Part 1 — Relation between $a$, $r$, $t$:

Let radiant energy $Q$ fall on a surface. Of this, let $Q_a$ be absorbed, $Q_r$ be reflected, and $Q_t$ be transmitted. By conservation of energy:

$Q = Q_a + Q_r + Q_t$

Dividing throughout by $Q$:

$1 = \frac{Q_a}{Q} + \frac{Q_r}{Q} + \frac{Q_t}{Q}$

Defining the coefficient of absorption $a = Q_a/Q$, coefficient of reflection $r = Q_r/Q$, and coefficient of transmission $t = Q_t/Q$ (each a pure number between 0 and 1):

$\boxed{a + r + t = 1}$

(For a perfectly black body, $a=1, r=t=0$.)

Part 2 — RMS speed comparison:

R.m.s. speed of a gas: $v_{rms} = \sqrt{\dfrac{3RT}{M}}$, where $M$ is the molar mass.

Hydrogen: $T_H = 127°\text{C} = 400\ \text{K}$, $M_H = 2\ \text{g/mol}$. …