1000 Solved Problems in Modern Physics

7.3 Solutions 417

Optical Doppler effect is given by

hν=hν 0 (1+βcosθ∗)

The maximum and minimum energy of photons will behν 0 (1+β) and

hν 0 (1−β)or661keV± 0 .5eV

7.84 The gravitational red shift is due to the change in the energy of a photon as
it moves from one region of space to another differing gravitational poten-
tial. The photon carries an inertial as well as the gravitational mass given by
hν/c^2. In its passage from a point where the gravitational potential isφ 1
to another point where the potential isφ 2 there will be expenditure of work
given byhν/c^2 times the potential difference (φ 2 −φ 1 ). This would result
in an equivalent decrease in the energy content of the photon and hence its
frequency.
ΔE=E(φ 2 −φ 1 )/c^2
A level difference ofHnear the earth’s surface would result in the fractional
shift of frequency
Δν/ν=gH/c
Now,Δν/ν=v/c=gH/c^2
orv=gH/c= 9. 8 × 22. 6 / 3 × 108 = 7. 38 × 10 −^7 m/s= 7. 38 × 10 −^4 mm/s
Thus resonance fluorescence would occur for downward velocity of the
absorber of magnitude 7.38 mm/s.

7.3.9 Radioactivity (General) .........................

7.85

T 0 4 8 1216202428323640

dN/dt 18.59 13.27 10.68 9.34 8.55 8.03 7.63 7.30 6.99 6.71 6.44

ln(dN/dt) 2.923 2.585 2.368 2.234 2.146 2.083 2.032 1.988 1.944 1.904 1.863

The log-linear plot of dN/dtversus time (t) is not a straight line because

the source contains two types of radioactive material of different half-lives

described by the sum of two exponentials. If the two half-lives are widely dif-

ferent then it is possible to estimate the half-lives by the following

procedure.

Toward the end of the curve (Fig. 7.18), say for time 20–28 h, most of the

atoms of the shorter-lived substance with half-lifeT 1 would have decayed

and the curve straightens up, corresponding to the single decay of longer

half-life ofT 2. If this straight line is extrapolated back up to they-axis, then

the half-lifeT 2 can be estimated in the usual way as from the slope of the

curve for a single source on the log-linear plot. In this example,

T 2 =

0. 693 Δt

ln

(dN

dt

)

0 −ln

(dN

dt

)

t

=

(0.693)(40)

(2. 083 − 1 .863)

=63 min

For the shorter-lived substance, the contributiondN 2 /dtof the source 2

can be subtracted from the observed values in the initial portion of the curve

over suitable time interval and the procedure repeated. In this way we find

T 1 =10 min.