6.105. For the state 4P: h;a2, h 15/2, and h V 32/2; for
the state 5D: 0, h1/ 2; h
h
h V 20.
6.106. (a) 2F7/2 , -
M max — hi/ 63/2, (b) 3F4, M 771aX = 2h V 6.
6.107. In the F state M8 = h ;-; for the D state it can be
only found that M,„>.-h . 1/ - 6-.
6.108. 3, 4, 5.
6.109. (a) 1, 3, 5, 7, 9; (b) 2, 4, 6; (c) 5, 7, 9.
6.110. 31°.
6.111. 3D 2.
6.112. 'PI, ip2, 11'3, 3P0,1,22 3D1,2,3, 3F2,3.4.
6.113. The same as in the foregoing problem.
6.114. The second and the third term.
6.115. g = 4 ± 6 = 10.
6.116. 4, 7 and 10.
6.117. 3F3.
6.118. As.
6.119. (a) 4S312; (b) 3P2.
6.120. (a) 4F312, W172; (b) 4F912, h3 1112.
6.121. (a) Two d electrons; (b) five p electrons; (c) five d electrons.
6.122. (a) 3P 0, (b) 4F912.
6.123. 4F312.
6.124. p. = to B Y35 ( 65512)•
6.125. 1 = n2e-0 /hT = 3.10-17, where o = R (1 — 1/n2).
6.126. NIN, = (g/g0) e-TiwIkT = 1.14.10-4, where g and go
are the statistical weights (degeneracy ratios) of the levels 3P and
3S respectively (g = 6, go = 2).
6.127. ti = 1111 In i = 1.3 1.ts.
6.128. N = XTP/2nch = 7.103.
6.129. r = (nhalP) (g/ go) e —hWAT = 65 ns, where g and go
are the degeneracy ratios of the resonant and the basic level.
6.130. (a) Pind/Po p = 1/(e")/hT — 1)
10-34,
where a) =
=-- 3/4R; (b) T = 1.7.105 K.
6.131. Suppose that I is the intensity of the passing ray. The
decrease in this value on passing through the layer of the substance
of thickness dx is equal to
— dI = xI dx = (N1B12 — N2B21) (Iic) ho) dx,
where N 1 and N2 are the concentrations of atoms on the lower and
upper levels, B12 and B21 are the Einstein coefficients. Hence
x a (hco/c) N1R12 (1 — g1N2ig2N1)•
Next, the Boltzmann distribution should be taken into consideration, as well as the fact that ho) >> kT (in this case N1 is approximately equal to No, the total concentration of the atoms).
6.132. AkDo p/Aknot
43T-rvp,./X,
103, where vp,. = 172RTIM.
357
the state 5D: 0, h1/ 2; h
h
h V 20.
6.106. (a) 2F7/2 , -
M max — hi/ 63/2, (b) 3F4, M 771aX = 2h V 6.
6.107. In the F state M8 = h ;-; for the D state it can be
only found that M,„>.-h . 1/ - 6-.
6.108. 3, 4, 5.
6.109. (a) 1, 3, 5, 7, 9; (b) 2, 4, 6; (c) 5, 7, 9.
6.110. 31°.
6.111. 3D 2.
6.112. 'PI, ip2, 11'3, 3P0,1,22 3D1,2,3, 3F2,3.4.
6.113. The same as in the foregoing problem.
6.114. The second and the third term.
6.115. g = 4 ± 6 = 10.
6.116. 4, 7 and 10.
6.117. 3F3.
6.118. As.
6.119. (a) 4S312; (b) 3P2.
6.120. (a) 4F312, W172; (b) 4F912, h3 1112.
6.121. (a) Two d electrons; (b) five p electrons; (c) five d electrons.
6.122. (a) 3P 0, (b) 4F912.
6.123. 4F312.
6.124. p. = to B Y35 ( 65512)•
6.125. 1 = n2e-0 /hT = 3.10-17, where o = R (1 — 1/n2).
6.126. NIN, = (g/g0) e-TiwIkT = 1.14.10-4, where g and go
are the statistical weights (degeneracy ratios) of the levels 3P and
3S respectively (g = 6, go = 2).
6.127. ti = 1111 In i = 1.3 1.ts.
6.128. N = XTP/2nch = 7.103.
6.129. r = (nhalP) (g/ go) e —hWAT = 65 ns, where g and go
are the degeneracy ratios of the resonant and the basic level.
6.130. (a) Pind/Po p = 1/(e")/hT — 1)
10-34,
where a) =
=-- 3/4R; (b) T = 1.7.105 K.
6.131. Suppose that I is the intensity of the passing ray. The
decrease in this value on passing through the layer of the substance
of thickness dx is equal to
— dI = xI dx = (N1B12 — N2B21) (Iic) ho) dx,
where N 1 and N2 are the concentrations of atoms on the lower and
upper levels, B12 and B21 are the Einstein coefficients. Hence
x a (hco/c) N1R12 (1 — g1N2ig2N1)•
Next, the Boltzmann distribution should be taken into consideration, as well as the fact that ho) >> kT (in this case N1 is approximately equal to No, the total concentration of the atoms).
6.132. AkDo p/Aknot
43T-rvp,./X,
103, where vp,. = 172RTIM.
357