Question

The de-Broglie wavelength of helium atom at a temperature of 89ŗC, will be

• A. 0.47 Å
• B. 0.58 Å
• C. 0.66 Å
• D. 0.73 Å

Answer 1

Answer: (C)

0.66 Å

Answer 2

$\lambda_{n} = \frac{25.6Å}{\sqrt{T}}$' l_He = $\frac{25.6Å}{2\sqrt{T}}$ = $\frac{25.6Å}{2 \times 19}$ = 0.66Å