Exercises for Chapter 8 part 1 – plus TT2 review 1.

Exercises for Chapter 8 part 1 – plus TT2 review
1. The hydrogen line spectrum is evidence for:
1
the Heisenberg Uncertainty Principle
2
the quantized nature of hydrogen atomic energy states
3
the fact that hydrogen is diatomic
4
the de Broglie equation
5
the Pauli Principle
1a. (extra) Which of the following statements are true (Fall 1999)?
(I) Ultraviolet light has a longer wavelength than infrared radiation.
(II) Blue light has more energy per photon than red light.
(III) X-rays travel much faster than radio waves.
(IV) The energy of a photon is inversely proportional to its frequency.
A) (I) and (II)
C) (II), (III), and (IV)
E) (II) only
B) (II) and (III)
D) (II) and (IV)
1
2. Which of the following has the shortest wavelength (Fall 2001)?
1
2
3
infrared radiation
ultraviolet radiation
gamma rays
High Energy
4
5
microwaves
green light
Low Energy
2
5. Calculate the wavelength of the photon emitted in the transition n
= 3 to n = 2 of the hydrogen atom. (This is the first line of the Balmer
series).
RH = 2.18  10 -18 J


1
1
 E  Z 2 RH  2  2 
n f 
 ni
Z = 1, ni = 2, nf = 3
h = 6.63 × 10 -34 J sec
1
1 1
 1
18
 E  RH  2  2   R H     0.303  10 J
3 
2
4 9
E
12

 457  10 Hz
h
1
3  10 m s
 
12
1
 457  10 s
c
8
Finally convert frequency to
wavelength = 656 nm
Already done in class notes
3
6. A ground state H atom absorbs a photon of radiation of
wavelength 102.6 nm. What is the principal quantum number
of the electron in the excited state?
4
12. X-ray photons of frequency 30.3 × 1016 Hz are used to remove
2s electrons from neon. The kinetic energy of the ejected electrons is
193.2 × 10-18 J. Calculate the minimum energy required to remove a
single 2s electron from neon.
2s electrons
X-rays
Ne
h
KE = h - h0
=
=
KE
=
h0
=
5
ch 8 #91 The work function is the energy that must be supplied to
cause the release of an electron from a material. The corresponding
photon frequency is the threshold frequency. The higher the energy of
the incident light, the more the kinetic energy of the emitted electrons.
The work function of mercury is 435 kJ/mol.
a) Can the photoelectric effect be obtained with mercury using
visible light?
h0 = 435 kJ/mol
Photons IN
electrons
emitted
metal surface
=
(divide by N)
0 =
(divide by h)

( = c / 0 )
=
KE = h - h0
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b) If 215 nm light I used, what is the kinetic energy of the photons
emitted?
215 nm
KE
 = c /  E = h
= h - h0
=
=
c) What is the velocity of the ejected electrons?
(watch units)
KE = ½ mv2
7
Paschen (IR)
n=
n=4
n=3
Balmer
Balmer
s (Vis)
n=2
Energy
Red
Green
Blue
Violet
ch 8 #96 Between which two orbits
of the Bohr hydrogen atom must an
electron fall to produce light of
wavelength 1876 nm
Lyman
series
(ultrav(UV)
Lyman
Series
n=1
8