1. Xtra Edge February 2012 - Career Point

3. A monochromatic light source of frequency
ν illuminates a metallic surface and ejects
photoelectrons. The photoelectrons having maximum
energy are just able to ionize the hydrogen atom in
ground state. When the whole experiment is repeated
with an incident radiation of frequency (5/6) ν, the
photoelectrons so emitted are able to excite the
hydrogen atom beam which then emits a radiation of
wavelength 1215 Å. Find the work function of the
metal and the frequency ν.
Sol. In the first case,
Emax = Ionization energy = 13.6 eV
= 21.76 × 10 –19 J
So, hν = 21.76 × 10 –19 J ....(1)
In the second case,
So,
hc
E'max =
λ
−34
6.
6×
10 × 3×
10
=
−10
1215×
10
=16.3×10 –19 J
5ν h
–19
= 16.3 × 10 + W ...(2)
6
Dividing Eq.(1) by Eq.(2)
−19
6 21.
76×
10 + W
=
5
−19
16.
3×
10 + W
Solving, we get
W = 11.0 × 10 – 19 J = 6.875 eV
−19
−19
21.
76×
10 + 11.
0×
10
From Eq.(1) ν =
−34
6.
6×
10
= 5 × 10 15 Hz
4. The radiation, emitted when an electron jumps from
n = 3 to n = 2 orbit in a hydrogen atom, falls on a
metal to produce photoelectrons. The electrons from
the metal surface with maximum kinetic energy are
made to move perpendicular to a magnetic field of
1/320 T in a radius of 10 –3 m. Find (i) the kinetic
energy of electrons, (ii) wavelength of radiation and
(iii) the work function of metal.
Sol. (i) Speed of an electron in the magnetic field,
Ber
v =
m
Kinetic energy of electrons
1 2
Emax = mv =
2
2
2
2
B e r
2m
2
8
−19
⎛ 1 ⎞ ( 1.
6×
10 ) × ( 10
= ⎜ ⎟⎠ ×
⎝ 320
−31
2×
9.
1×
10
= 1.374 × 10 –19 J
= 0.8588 eV
2
−3
)
2
(ii) Energy of the photon emitted from a hydrogen
atom
hc ⎡ 1 1 ⎤
hν = =
λ
⎢ −
2 2 ⎥
⎣2
3 ⎦
= 1.888 eV
Wavelength of radiation,
XtraEdge for IIT-JEE 24 FEBRUARY 2012
−34
6.
62×
10 × 3×
λ =
1.
888×
1.
6×
10
= 6.572 × 10 –7 m
= 6572 Å
8
10
−19
(iii) Work function of metal W = hν – Emax
= 1.8888 – 0.8588
= 1.03 eV
5. X-rays are produced in an X-ray tube by electrons
accelerated through a potential difference of 50.0 kV.
An electron makes three collisions in the target
before coming to rest and loses half of its kinetic
energy in each of the first two collisions. Determine
the wavelengths of the resulting photons. Neglect the
recoil of the heavy target atoms.
Sol. Initial kinetic energy of the electron = 50.0 keV
Kinetic energy after first collision = 25.0 keV
Energy of the photon produced in the first collision,
E1 = 50.0 – 25.0 = 25.0 keV
Wavelength of this photon
−34
8
hc 6.
6×
10 × 3×
10
λ1 = =
E
−19
3
1 1.
6×
10 × 25.
0×
10
= 0.495 × 10 –10 m = 0.495 Å
Kinetic energy of the electron after second collision
= 12.5 eV
Energy of the photon produced in the second
collision, E2 = 25.0 – 12.5 = 12.5 keV
Wavelength of this photon
hc
=
E2
6.
6 × 10
λ2 =
1.
6 × 10
= 0.99 × 10 –10 m
= 0.99 Å
−34
−19
8
× 3×
10
3
× 12.
5×
10
Kinetic energy of the electron after third collision = 0
Energy of the photon produced in the third collision,
E3 = 12.5 – 0 = 12.5 keV
This is same as E2. Therefore, wavelength of this
photon, λ3 = λ2 = 0.99 Å.