Question

Calculate the wavelength of radiation, which excites the electron of hydrogen atom from ground state to fourth energy level. Ionisation energy of hydrogen atom is 1312 kJ/mol

Answer 1

Answer : The wavelength of radiation which excites the electron from ground state to fourth energy level is 97.3 nm

Explanation :

The problem can be solved using Rydberg's formula for wavelength which is given below.

$\frac{1}{\lambda}= R_{h}[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}]$

Here n1 < n2

The electron jumps from ground state for which n1 = 1 to fourth energy level which is n2 = 4.

The Rydberg's constant for wavelength is 1.097 x 10⁷ m⁻¹.

Let us plug in the above values in the given formula.

$\frac{1}{\lambda}= 1.097 \times 10^{7}m^{-1}[\frac{1}{1^{2}}-\frac{1}{4^{2}}]$

$\frac{1}{\lambda}= 1.097 \times 10^{7}m^{-1}[1-0.0625]$

$\frac{1}{\lambda}= 1.097 \times 10^{7}m^{-1}(0.9375)$

$\frac{1}{\lambda}= 1.028 \times 10^7 m^{-1}$

$\lambda= \frac{1}{1.028 \times 10^7 m^{-1}}$

$\lambda= 9.73 \times 10^-8 m$

The wavelength of radiation which excites the electron from ground state to fourth energy level is 97.3 nm or 9.73 x 10⁻8 m

Answer 2

Answer: Wavelength of the radiation is 97.29nm.

Explanation: Ionization energy is defined as the measure of energy from (+1) energy level to (infinity) energy level.

Mathematically,

$I.E.=E_{\infty }-E_1$

Energy associated to the $n^{th}$- energy level will be:

$E_n=\frac{E_o\times Z^2}{n^2}$     ...(1)

For ground state energy level of H-atom, n = 1 and Z = 1

Ionization energy = 1312 kJ/mol (given)

Putting the values in equation 1, we get

$1312kJ/mol=0-\frac{E_o\times 1^2}{1^2}$

$E_o=-1312kJ/mol$

Now, to calculate the wavelength of the radiation, we need to calculate the energy difference between fourth energy level and ground energy level.

$\Delta E=E_4-E_1$   ....(2)

Using Planck's equation,

$\Delta E=\frac{N_Ahc}{\lambda }$

Where, $N_A=6.022\times 10^{23}atoms/mol$ (Avogadro's number)

$h=6.624\times 10^{-34}Js$ (Planck's constant)

$c=3\times 10^8ms^{-1}$ (Speed of light)

Putting this equation in equation 2, we get

$\frac{N_Ahc}{\lambda}=\frac{E_o\times 1^2}{4^2}-\frac{E_o\times 1^2}{1^2}$

$\frac{(6.022\times 10^{23})(6.624\times 10^{-34})(3\times 10^8)}{\lambda}=\left (\frac{-1312\times 1^2}{4^2}  \right )-\left (\frac{-1312\times 1^1}{1^1}  \right )$

$\lambda =0.09723\times 10^{-6}m=97.29nm$ (Conversion Factor: $1m=1\times 10^{-9}nm$)

Wavelength of the radiation is 97.29 nm.