Question

A proton captures a free-electron whose K.E. is zero & forms a hydrogen atom of lowest energy level (n=1). If a photon is emitted in this process, then the wavelength of radiation and the region of electromagnetic spectrum, in which radiation fall is:

[Ionisation potential of hydrogen = 13.6 volt,$h=6.6\times {10}^{-34}K/s,C=3.0\times {10}^{8}m/s]$

• A. $910\stackrel{o}{A};\text{U.V.}$
• B. $230\stackrel{o}{A};\text{I.R.}$
• C. $450\stackrel{o}{A};\text{visible}$
• D. $980\stackrel{o}{A};\text{I.R}$

Answer 1

The correct option is A $910\stackrel{o}{A};\text{U.V.}$

The ionisation energy is given by

$IE=E_{\infty }-E_1=\Delta E$

$13.6=0-E_1$

$E_1=-13.6eV\left(\text{energy in ground state}\right)$

$\frac{hc}{\lambda }=\Delta E$

$\Rightarrow \lambda =\frac{hc}{\Delta E}=\frac{1242eV.nm}{13.6eV}=91nm=910A^o$

It falls in ultraviolet region.