Question

What is the frequency and wavelength of a photon during transition from $n = 5\,\,to\,\,n = 2$ state in the $H{e^ + }\,ion$.

Answer 1

We need to learn the atomic number of the elements in the periodic table. Hence atomic number of $He^+$ is represented as $Z$ i.e. $Z = 2$.Then by using a single electron species formula we can calculate the frequency and wavelength where ${n_1} = 2\,\,and\,\,{n_2} = 5$.

Formula Used:
$\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
Where,
$\lambda =$ Wavelength
$R =$ Rydberg Constant
$n =$ Principle Quantum number
$v = \dfrac{c}{\lambda }$
Where,
$v =$ Frequency
$c =$ Speed of light

Complete step by step solution:
A photon transmits from $n = 5\,\,to\,\,n = 2$ state. While transition some wavelength and frequency is generated which we have to calculate.
As given in the question,
A photon in $H{e^ + }\,\,ion$ state moves from ${n_2} = 5\,\,to\,\,{n_1} = 2$ .
Using Single Electron Species, we get
$\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right) \cdot \cdot \cdot \cdot \left( 1 \right)$
Where,
$\lambda =$ Wavelength
$R =$ Rydberg Constant
$\Rightarrow R = 1.097 \times {10^7}\,{m^{ - 1}}$
$Z =$ Atomic number of $H{e^ + }$ ion $= 2$
${n_1}$ and ${n_2}$ (Given)
Putting all these values in the equation $\left( 1 \right)$ , we get
$\dfrac{1}{\lambda } = 1.097 \times {10^7}{m^{ - 1}} \times {2^2}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{5^2}}}} \right)$
As Wavelength is calculated in $nm$convert the above equation into $nm$ by multiplying the above formula with ${10^{ - 9}}$ .
$\dfrac{1}{\lambda } = 1.097 \times {10^7} \times {10^{ - 9}}n{m^{ - 1}} \times 4\left( {\dfrac{1}{4} - \dfrac{1}{{25}}} \right)$
$\Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^{ - 2}}n{m^{ - 1}} \times 4\left( {\dfrac{{25 - 4}}{{100}}} \right)$
$\Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^{ - 2}}n{m^{ - 1}} \times \left( {\dfrac{{21}}{{25}}} \right)$
$\Rightarrow \dfrac{1}{\lambda } = \dfrac{{23.037 \times {{10}^{ - 2}}}}{{25}}n{m^{ - 1}}$
$\Rightarrow \lambda = \dfrac{{25 \times 100}}{{23.037}}nm$
$\Rightarrow \lambda = 108.5nm$
Hence the wavelength a photon generates during transition is $108.5nm$ .
Now we have to calculate the frequency,
We know that,
$v = \dfrac{c}{\lambda }$
Here
$c = 3.8 \times {10^8}m{s^{ - 1}}$ -Speed of light
By putting the respective values in the frequency equation we will get,
$v = \dfrac{{3.8 \times {{10}^8}m{s^{ - 1}}}}{{108.5nm}}$
Now to get a simplified value of frequency we need to convert $nm$ into $m$ .
Now multiply the denominator with ${10^{ - 9}}$.
We get,
$v = \dfrac{{3.8 \times {{10}^8}m{s^{ - 1}}}}{{108.5 \times {{10}^{ - 9}}m}}$
$\Rightarrow v = 0.035 \times {10^{8 + 9}}{s^{ - 1}}$
$\Rightarrow v = 0.035 \times {10^{17}}{s^{ - 1}}$
$\Rightarrow v = 3.50 \times {10^{15}}{s^{ - 1}}$
Therefore the frequency a photon generates during transition is $3.50 \times {10^{15}}{s^{ - 1}}$.

Note:
We always keep in mind while calculating wavelengths that convert the meter into nanometers. Generally we make a common mistake while writing the atomic number of an ion. Atomic number never depends on the number of electrons, it depends on the number of protons present in an element.