Question

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

Answer 1

Hint: We need to learn the atomic number of the elements in the periodic table. Hence atomic number of $H{e}^{+}$ is represented as $Z$ i.e. $Z=2$.Then by using a single electron species formula we can calculate the frequency and wavelength where $n_1=2and{n}_2=5$.

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


Complete step by step solution:
A photon transmits from $n=5ton=2$ 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 $n_2=5to{n}_1=2$ .
Using Single Electron Species, we get
${\displaystyle \frac{1}{\lambda }}=R{Z}^2\left({\displaystyle \frac{1}{{n_1}^2}}-{\displaystyle \frac{1}{{n_2}^2}}\right)\cdot \cdot \cdot \cdot \left(1\right)$
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
${\displaystyle \frac{1}{\lambda }}=1.097\times {10}^7{m}^{-1}\times 2^2\left({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{5^2}}\right)$
As Wavelength is calculated in $nm$convert the above equation into $nm$ by multiplying the above formula with ${10}^{-9}$ .
${\displaystyle \frac{1}{\lambda }}=1.097\times {10}^7\times {10}^{-9}n{m}^{-1}\times 4\left({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{25}}\right)$
$\Rightarrow {\displaystyle \frac{1}{\lambda }}=1.097\times {10}^{-2}n{m}^{-1}\times 4\left({\displaystyle \frac{25-4}{100}}\right)$
$\Rightarrow {\displaystyle \frac{1}{\lambda }}=1.097\times {10}^{-2}n{m}^{-1}\times \left({\displaystyle \frac{21}{25}}\right)$
$\Rightarrow {\displaystyle \frac{1}{\lambda }}={\displaystyle \frac{23.037\times {10}^{-2}}{25}}n{m}^{-1}$
$\Rightarrow \lambda ={\displaystyle \frac{25\times 100}{23.037}}nm$
$\Rightarrow \lambda =108.5nm$
Hence the wavelength a photon generates during transition is $108.5nm$ .
Now we have to calculate the frequency,
We know that,
$v={\displaystyle \frac{c}{\lambda }}$
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,
$v={\displaystyle \frac{3.8\times {10}^{8}m{s}^{-1}}{108.5nm}}$
Now to get a simplified value of frequency we need to convert $nm$ into $m$ .
Now multiply the denominator with ${10}^{-9}$.
We get,
$v={\displaystyle \frac{3.8\times {10}^{8}m{s}^{-1}}{108.5\times {10}^{-9}m}}$
$\Rightarrow v=0.035\times {10}^{8+9}{s}^{-1}$
$\Rightarrow v=0.035\times {10}^{17}{s}^{-1}$
$\Rightarrow v=3.50\times {10}^{15}{s}^{-1}$
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.