Question

The wavelength of light emitted in the visible region by $H{e}^{+}$ ions after collisions with $H$ atoms is:

• A. $6.5\times {10}^{-7}m$
• B. $5.6\times {10}^{-7}m$
• C. $4.8\times {10}^{-7}m$
• D. $4.0\times {10}^{-7}m$

Answer 1

The correct option is C $4.8\times {10}^{-7}m$

	$H-$ atom	$H{e}^{+}$atom
$n=1$	$-13.6$ eV	$-54.4$ eV
$n=2$	$-3.4$ eV	$-13.6$ eV
$n=3$	$-1.51$ eV	$-6.04$ eV
$n=4$	$-0.85$ eV	$-3.4$ eV

Energy transfer by $H$ atom to $H{e}^{+}$ atom is 10.2 eV. Subsequently $H{e}^{+}$ gets excited to $n=4$.

Only the energy corresponding to transition for $n=4$ to $n=3$ falls in the visible region.

So, $\lambda =\frac{hc}{2.64eV}=4.8\times {10}^{-7}m$