Question

For the hydrogen atom, the transition from the 2p state to the 1s state is accompanied by the emission of a photon with an energy of 16.2x10-19 J. For a Ni atom, the same transition (2p to 1s) is accompanied by the emission of x-rays of wavelength 1.66 A. What is the energy difference between these states in nickel? The energy difference between these states in nickel is the corresponding energy difference for hydrogen The energy difference between these states in selenium would be expected to be the corresponding energy difference for nickel.

Answer 1

The energy difference between the two states ΔΕnickel=hc/λ

where h = Planck's constant = 6.6*10^(-34) Js;

c= 3*10^8 m/s &

λ = wavelength = 1.66*10^(-10) m.

ΔΕnickel = 1.19 * 10^(-15) J.

This is (1.19*10^(-15) / 16.2*10^(-19)) = 734.56 times than that of H-atom.

The energy difference between any two states is given by

ΔE = -13.6(Z^2){(1/nf^2) - (1/ni^2)}

Here, Z = nuclear charge.

nf = The principal quantum number of the final state.

ni = The principal quantum number of the initial state.

For Se, ΔΕselenium = {(Zselenium/Znickel)^2}*ΔEnickel.

Zselenium = 34

Znickel = 28

Therefore ΔEselenium = {(34/28)^2} times that of ΔEnickel.

= 1.4745 times that of ΔΕnickel.