Question

a) Write the electronic configuration of Cu in s p d f

b) Which of the following orbitals has the lowest energy
4d, 4f, 5s, 5p

c) What is the wavelength of light emitted when the electrons in a hydrogen atom undergo the transition from an energy level n=4 to an energy level n=2?

Answer 1

$\left(a\right)$ Electronic configuration of $Cu=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^1$

$\left(b\right)$

$4d$ $n+1=4+2=6$

$4f$ $n+1=4+3=7$

$5s$ $n+1=5+0=5$

$5p$ $n+1=5+1=6$

So, $5s$ has the lowest energy.

$\left(c\right)$

$\frac{1}{\lambda }=R_H(\frac{1}{h_1^2}-\frac{1}{h_2^2})\dots$

$\frac{1}{\lambda }=R_H(\frac{1}{2^2}-\frac{1}{4^2})$

$\frac{1}{\lambda }=R_H(\frac{1}{4}-\frac{1}{16})$

$\lambda =\frac{16}{3R_H}$

$=\frac{16}{3}\times 91.6$ Since, ( $\frac{1}{R_H}=91.6nm$ )

$=488.5nm$