Chapter 3, Problem 3.123QA

Interpretation Introduction

To find:

The transaction energies from different energy levels in the helium ions

Answer to Problem 3.123QA

Solution:

a) The statement is true

b) The statement is false

c) The statement is true

d) The statement is true

Explanation of Solution

1) Concept:

We are asked to determine the energy associated with electron transactions.

We need to identify the initial and final energy levels and calculate the transaction energy by using Bohr’s energy formula.

2) Formula:

i. $∆E= -2.178 \times {10}^{-18} J \times Z^2 \left(\frac{1}{{n_{final}}^2}-\frac{1}{{n_{initial}}^2} \right)$

ii. $\lambda = \frac{hc}{∆E}$

iii. $\nu = \frac{c}{\lambda }$

3) Calculations:

Apply the Rydberg energy formula using Z = 2 for He⁺ ion, n =3 to n= 1

$∆E= -2.178 \times {10}^{-18} J \times Z^2 \left(\frac{1}{{n_{final}}^2}-\frac{1}{{n_{initial}}^2} \right)$

$∆E= -2.178 \times {10}^{-18} J \times 2^2 \left(\frac{1}{1^2}-\frac{1}{3^2} \right)$

$∆E= -7.74 \times {10}^{-18} J$

a) The transaction energy for n=3 to n= 1 is -7.74 x 10^-18 J.

Energy lost from n= 3 to n = 2 is

$∆E= -2.178 \times {10}^{-18} J \times 2^2 \left(\frac{1}{2^2}-\frac{1}{3^2} \right)$

$∆E_1= -1.21 \times {10}^{-18} J$

Energy lost from n= 2 to n = 1 is

$∆E= -2.178 \times {10}^{-18} J \times 2^2 \left(\frac{1}{1^2}-\frac{1}{2^2} \right)$

$∆E_2= -6.53 \times {10}^{-18} J$

Total energy between two step process is

$∆E= ∆E_1+ ∆E_2$

$∆E=\left(-1.21 \times {10}^{-18} J\right)+\left(-6.53 \times {10}^{-18} J\right)= -7.74 \times {10}^{-18} J$

So the total energy is same as the energy lost between two step process.

b) We will calculate the wavelength for n =3 to n= 1.

$\lambda = \frac{hc}{∆E}$

$\lambda = \frac{6.626\times {10}^{-34} j.s \times 2.998 \times {10}^8 m/s}{7.74 \times {10}^{-18} J}$

$\lambda =2.57 \times {10}^{-8} m$

Calculate the wavelength for n =3 to n= 2.

$\lambda = \frac{6.626\times {10}^{-34} j.s \times 2.998 \times {10}^8 m/s}{6.53 \times {10}^{-18} J}$

$\lambda =3.04 \times {10}^{-8} m$

Calculate the wavelength for n =2 to n= 1.

$\lambda = \frac{6.626\times {10}^{-34} j.s \times 2.998 \times {10}^8 m/s}{1.21 \times {10}^{-18} J}$

$\lambda =1.64 \times {10}^{-7} m$

Total wavelength for two process is

$\lambda =3.04 \times {10}^{-8} m+ 1.64 \times {10}^{-7} m =1.94 \times {10}^{-7} m$

The wavelength from n= 3 to n= 1 is not same with sum of two process.

c) The formula of frequency is

$\nu = \frac{c}{\lambda }$

First calculate the frequency for n= 3 to n =1 transaction.

$\nu = \frac{2.998 \times {10}^8 m/s}{2.57 \times {10}^{-8} m}=1.17 \times {10}^{16} s^{-1}$

Frequency for n= 3 to n =2

$\nu = \frac{2.998 \times {10}^8 m/s}{2.57 \times {10}^{-8} m}=1.83 \times {10}^{15} s^{-1}$

Frequency for n= 2 to n =1

$\nu = \frac{2.998 \times {10}^8 m/s}{2.57 \times {10}^{-8} m}=9.86 \times {10}^{15} s^{-1}$

The sum of two transaction frequency is

$\lambda =1.83 \times {10}^{15} s^{-1}+ 9.86 \times {10}^{15} s^{-1} =1.17 \times {10}^{16} s^{-1}$

So sum of frequency for two photons emitted is equals to the frequency emitted for single photon.

d) The wavelength for He⁺ ion is

$\lambda = \frac{hc}{∆E}$

$\lambda = \frac{6.626\times {10}^{-34} j.s \times 2.998 \times {10}^8 m/s}{7.74 \times {10}^{-18} J}= 2.57 \times {10}^{-8} m$

The energy and wavelength of H atom are

$∆E= -2.178 \times {10}^{-18} J \left(\frac{1}{1^2}-\frac{1}{3^2} \right)= -1.94 \times {10}^{-18} J$

$\lambda = \frac{6.626\times {10}^{-34} j.s \times 2.998 \times {10}^8 m/s}{1.94 \times {10}^{-18} J}=1.03 \times {10}^{-7} m$

The wavelength for He⁺ ions is shorter than H atom.

Conclusion

We used Bohr’s energy relation for calculation of energy, wavelength and frequency.