Answer
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Hint: 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.
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.
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