Question

If the ionization energy of hydrogen atom is $13.6eV$ then the wavelength of the radiation required to excite the electron in $L{i}^{++}$ from first to third Bohr orbit is approximately

• A. $1140A$
• B. $914A$
• C. $11.4A$
• D. $134A$

Answer 1

The correct option is B $914A$

Ionisation energy is given by

$E=13.6eV$ (given)

or $E=13.6\times 1.6\times {10}^{-19}J$ ....(1)

also $E=hv$

$E=\frac{hc}{\lambda }$ ....(2)

equation (1) and (2)

$\frac{hc}{\lambda }=13.6\times 1.6\times {10}^{-19}$

$\lambda =\frac{h\times c}{13.6\times 1.6\times {10}^{-19}}$

$\lambda =\frac{3\times {10}^8\times 6.63\times {10}^{-34}}{13.6\times 1.6\times {10}^{-19}}$

$\lambda =914\times {10}^{-10}m$

$\lambda =914A^o$