# BOHR MODEL & X-RAYS - 4

__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 10^{7}(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 10^{18} 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 10^{7})(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