Question

Ultraviolet light of wavelength ${\lambda }_1$ and ${\lambda }_2$ when allowed to fall on hydrogen atoms in their ground state is found to liberate electron with kinetic energy $1.8eV$ and $4.0eV$ respectively. Find the value of $\frac{{\lambda }_1}{{\lambda }_2}$.

• A. $\frac{7}{8}$
• B. $\frac{8}{7}$
• C. $\frac{9}{20}$
• D. $\frac{20}{9}$

Answer 1

The correct option is D $\frac{20}{9}$

We know that, energy and wavelength is related as,

$E=hv=\frac{hc}{\lambda }\{\because v\lambda =c\}$

$\therefore$ Here,

Hence, $E_1=1.8$ $eV$

$\Rightarrow 1.8=\frac{hc}{{\lambda }_1}$.... equation $01$

Also, $E_2=4$ $eV$

$\Rightarrow 4=\frac{hc}{{\lambda }_2}$...... equation $02$

Dividing equation $02$ and $01$, we get

$\frac{hc}{{\lambda }_2}÷\frac{hc}{{\lambda }_1}=\frac{4}{1.8}$

$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\frac{4}{1.8}=\frac{20}{9}$