Question

Consider a hypothetical hydrogen like atom. The wavelength in $A$ for the spectral lines for transition from $n=p$ to $n=1$ are given by-$\lambda =\frac{1500p^2}{p^2-1}$
Where $p=2,3,4,.$ (given $hc=12400eV/A$)

• A. The wavelength of the least energetic and the most energetic photons in this series is $2000A$, $1500A$
• B. Difference between energies of fourth and this orbit is $0.40eV$.
• C. Energy of second orbit is $6.2eV$
• D. The ionization potential of this element is $8.27V$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A The wavelength of the least energetic and the most energetic photons in this series is $2000A$, $1500A$
B Difference between energies of fourth and this orbit is $0.40eV$.
C Energy of second orbit is $6.2eV$
D The ionization potential of this element is $8.27V$

Given,

$\lambda =\frac{1500p^2}{p^2-1}=\frac{1500}{1-\frac{1}{p^2}}{A}^{\cdot }$

${\lambda }_{max}$ at p=2 is $2000A^{\cdot }$

${\lambda }_{min}$ at p=$\infty$ is $1500A^{\cdot }$

$E_4=\frac{12400}{1500}(1-\frac{1}{p^2})=\frac{124}{15}\times \frac{15}{16}$

$E_3=\frac{124}{15}(1-\frac{1}{9})$ $=\frac{124}{15}\times \frac{8}{9}$

$E_2=\frac{124}{15}(1-\frac{1}{4})$ $=\frac{124}{15}\times \frac{3}{4}=\text{6.2 eV}$

Now,

$E_4-E_3=\frac{124}{15}\times (\frac{15}{16}-\frac{8}{9})=\frac{124}{15}\times \frac{7}{144}=\frac{868}{144\times 15}=\text{0.40 eV}$

and, $E_{\infty }=\text{Ionisation Energy}$

$\text{Ionisation Energy}=-E_{\infty }=-\frac{124}{15}=-8.27eV$

$\text{Ionisation Potential}=\frac{\text{Ionisation energy}}{-e}$

$\text{Ionisation Potential}=8.27V$

Hence, Options $\text{A,B,C,D}$ are correct.