Homework Solutions

January 5

22.    l = 7.80 x 10² nm = 7.80 x 10^-7 m   
c = 2.99792458 x 10⁸ m s^-1
l = c/n       n = c/l = 3.84 x 10¹⁴ s^-1

E = hn

E = 2.55 x 10^-19 J

36.    l = h/mv            Note:  1 J = 1 kg m²s^-2

a.    m_p = 1.67262 x 10^-27 kg, v_p = 0.15*c, l = 8.81 x 10^-15 m

b.    m_e = 9.10939 x 10^-31 kg, ve = 0.15*c, l = 1.62 x 10^-11 m = 0.162 angstrom (atom diameter is about 1 angstrom)

c.    m_ball = 0.150 kg, v_ball = 10. m/s, l = 4.42 x 10^-34 m

38.    l = 1.0 x 10² nm = 1.0 x 10^-7 m and l = 1.0 x 10^-9 m;

v = h/lm; v = 7274 m/s and v = 72.74 m/s

53.    Because--according to theory--the wave goes on into infinity.  And we want to restrain our concern to the most relevant portion of the orbital; that is, the region closest to the nucleus.

58. The probability that the electron is at that point.