Question

The energy level diagram of the given element is given below. Identify by doing necessary calculation which transition corresponds to the emission of a spectral line of wavelength 102.7 nm.

Answer 1

Hint: The amount of energy associated with the photon of radiation is directly proportional to the frequency of incident light which is expressed as $E=h\nu$. It shows that higher the frequency of incident light the higher the energy corresponds to photons.

Complete answer:
As we know the relation between energy of photon and frequency of incident light is $E=h\nu$.
Where is defined as a universal constant, known as planck's constant. Value of planck's constant is $6.626\times {10}^{-34}$ Js.
 As we know frequency is also expressed in terms of light constant and wavelength of radiation.
$\nu ={\displaystyle \frac{c}{\lambda }}$
So, final equation of energy in terms of wavelength becomes-
$E={\displaystyle \frac{hc}{\lambda }}$
For element A
$E_1=-1.5$
$E_2=-0.85$
Therefore, difference in energy is
$\mathrm{\Delta }\mathrm{E}=\left(E_2-E_1\right)$
$\mathrm{\Delta }\mathrm{E}=-0.85-\left(-1.5\right)$
After solving the above equation, we get $\mathrm{\Delta }\mathrm{E}=0.65$
Now calculate the wavelength of transition
$\lambda ={\displaystyle \frac{hc}{E}}$
$\lambda ={\displaystyle \frac{6.626\times {10}^{-34}\times 3.0\times {10}^8}{0.65\times 1.6\times {10}^{-19}}}$
After solving this above equation, we get
$\lambda =19.038\times {10}^{-7}m$
In terms of nanometer value of wavelength is-
$\lambda =1903.8nm$
For element B
$E_1=-3.4$
$E_2=-0.85$
Therefore, difference in energy is
$\mathrm{\Delta }\mathrm{E}=\left(E_2-E_1\right)$
$\mathrm{\Delta }\mathrm{E}=-0.85-\left(-3.4\right)$
After solving the above equation, we get $\mathrm{\Delta }\mathrm{E}=2.55$
Now calculate the wavelength of transition
$\lambda ={\displaystyle \frac{hc}{E}}$
$\lambda ={\displaystyle \frac{6.626\times {10}^{-34}\times 3.0\times {10}^8}{2.55\times 1.6\times {10}^{-19}}}$
After solving this above equation, we get
$\lambda =4.8529\times {10}^{-7}m$
In terms of nanometer value of wavelength is-
$\lambda =485.2nm$
For element C
$E_1=-3.4$
$E_2=-1.5$
Therefore, difference in energy is
$\mathrm{\Delta }\mathrm{E}=\left(E_2-E_1\right)$
$\mathrm{\Delta }\mathrm{E}=-1.5-\left(-3.4\right)$
After solving the above equation, we get $\mathrm{\Delta }\mathrm{E}=1.9$
Now calculate the wavelength of transition
$\lambda ={\displaystyle \frac{hc}{E}}$
$\lambda ={\displaystyle \frac{6.626\times {10}^{-34}\times 3.0\times {10}^8}{1.9\times 1.6\times {10}^{-19}}}$
After solving this above equation, we get
$\lambda =6.5131\times {10}^{-7}m$
In terms of nanometer value of wavelength is-
$\lambda =651nm$
For element D
$E_1=-13.6$
$E_2=-1.5$
Therefore, difference in energy is
$\mathrm{\Delta }\mathrm{E}=\left(E_2-E_1\right)$
$\mathrm{\Delta }\mathrm{E}=-1.5-\left(-13.6\right)$
After solving the above equation, we get $\mathrm{\Delta }\mathrm{E}=12.1$
Now calculate the wavelength of transition
$\lambda ={\displaystyle \frac{hc}{E}}$
$\lambda ={\displaystyle \frac{6.626\times {10}^{-34}\times 3.0\times {10}^8}{12.1\times 1.6\times {10}^{-19}}}$
After solving this above equation, we get
$\lambda =1.0227\times {10}^{-7}m$
In terms of nanometer value of wavelength is-
$\lambda =102.27nm$
Hence, from the above calculation transition of element D from $-1.5$ ev to $-13.6$ ev corresponds with the emission of wavelength of $102.27nm$.

Note:
Make sure to convert the wavelength of the radiation into nanometers and convert the unit of $\mathrm{\Delta }\mathrm{E}$ from volts to electron-volts by multiplying it by $1.6\times {10}^{-19}$.