MOSELEY'S LAW FOR CHARACTERISTIC X-RAY:-
Moseley measured the frequency of characteristic X-rays from a large number of elements and plotted the Vs Z. The plot was close to a straight line (emission was for K_d line).
Moseley forwarded his law of the frequency of K_d lin:-
= a(Z - 1) = 4.98 x 10⁷(Z - 1)
where 'a' is a constant
for general characteristic X-rays, Moseley's law is
= a(Z - b) b is another constant.

ILLUSTRATION:- Use Moseley's law with b = 1 to find the frequency of the K_d X-rays of La(Z = 57) if the frequency of K_d X-rays of Cu(Z = 2a) is known to be 1.88 x 10¹⁸ Hz.
Solution:- Using equation = a(Z - b)



PROBLEM:-
Easy:- 1. An experiment measuring the K_d line of Fe yields 1.94 . Determine atomic number of iron.
Solution:- From Moseley law
= (4.97 X 10⁷)(Z - 1) or Z = i +
also f =
We get Z = 1 +
Z = 260.2 26 (atomic number of iron)

2. Stopping potential of 24, 100, 110 and KV are measured for electrons emitted from o certain element when it is irradiated with monochromatic X-ray. If this element is used as target in an X-ray tube, What will be the waavelength of the line.
Solution:- The stopping energy, eV_s is equal to the difference between the energy of the incident photon and the binding energy of the electron in a particular shell.
eV_s = E_P - E_B
The different stopping potential arise from electrons being emitted from different shells, with the smallest value (24 KV) corresponding to ejection of a K - shell electron. Subtrating the expression for the two smallest stopping potentials, we obtain
eV_SL - eV_SK = (E_P - E_BL) - (E_P - E_BK) = E_BK - E_BL
or 100 KeV - 24 KeV = E_BK - E_BL
The difference, 76 KeV is the energy of the K_d line. The corresponding wavelength is
= 0.163

3. The wavelength of the first member of the balmen series in hydrogen spectrum is 6563 . What is the wavelength of the first member of Lymen series ?
Solution:- Balmen series
Lymen series



4. The wavelength of K_d X-rays tungsten is 0.21 . If the enrgy of tungsten atom with an L electron knocked out in 11.3 KeV. What will be the energy of this atom when a K electron is knocked out ?
Solution:- Energy of K_d photon : E =
=> E = = 59.1 KeV
Let E_K = energy of the with a vacancy in the K - shell
E_L = energy of the atom with vacancy in the L - shell.
Then E_K - E_L = E => E_K + E_L = 59.1 + 11.3 = 70.4 KeV