Question

Consider a hypothetical hydrogen like atom $\left(\text{Ground state energy = 0}\right)$. The wavelength in $\mathring{\text{A}}$ for the spectral lines for transition from $n=p$ to $n=1$ are given by:
$\lambda =\frac{1500p^2}{p^2-1}$ where $p=2,3,4,...$

$(\text{Take}hc=12400\text{eV}\mathring{\text{A}})$

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

Answer 1

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

The ionization of this element is $8.27\text{eV}$

Given that,

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

$\left(A\right)$ The least energetic photon corresponds to the transition from $n=2\to n=1$.

So, substituting $p=2$ ,

$\Rightarrow {\lambda }_{max}=\frac{1500}{1-\frac{1}{2^2}}=\frac{4}{3}\times 1500\mathring{\text{A}}$

$\Rightarrow {\lambda }_{max}=2000\mathring{\text{A}}$

The most energetic photon corresponds to the transition from $n=\infty \to n=1$. So, $p=\infty$,

$\Rightarrow {\lambda }_{min}=\frac{1500}{1-\frac{1}{{\infty }^2}}=1500\mathring{\text{A}}$

Now,

${\lambda }_{2\to 1}=\frac{1500}{1-\frac{1}{4}}=2000\mathring{\text{A}}$

$\Rightarrow \Delta E_{2\to 1}=\frac{hc}{{\lambda }_{2\to 1}}=\frac{12400}{2000}\text{eV}=6.2\text{eV}$

${\lambda }_{3\to 1}=\frac{1500}{1-\frac{1}{9}}=\frac{9}{8}\times 1500\mathring{\text{A}}$

$\Rightarrow \Delta E_{3\to 1}=\frac{12400}{\frac{9}{8}\times 1500}\text{eV}=7.35\text{eV}$

Simlarly,

$\Delta E_{4\to 1}=\frac{12400}{\frac{16}{15}\times 1500}\text{eV}=7.75\text{eV}$

Given that $E_1=0$, From the above data,

$E_2=6.2\text{eV}$
$E_3=7.35\text{eV}$
$E_4=7.75\text{eV}$

$\Rightarrow E_4-E_3=0.40\text{eV}$

So, option $B\text{and}C$ are correct.

The ionization energy is the energy required to take an electron from $n=1\to n=\infty$

$E_{ion}=\frac{hc}{{\lambda }_{min}}=\frac{12400}{1500}\text{eV}=8.27\text{eV}$

Hence, all the given options are correct.