Electrons are accelerated through an electric potential V and then fall on a pair of slits that have a separation of 100 nm. The resultant interference pattern indicates that the electrons have a wavelength of 1 nm. i) What is the value of the accelerating electric potential V? ii) After passing through the slits what is the minimum spread in the electron’s momentum in the direction parallel to the plane of the slits and perpendicular to the average path of the electrons? Assume ℎ = 6.626 × 10−34 J.s and electron mass, m = 9.11 × 10−31 kg. Ignore any relativistic effect.
a) i) 1 V ii) infinity
b) i) 1 V ii) zero
c) i) 1.5 V ii) 10.54e-28 kg.m/s
d) None of these
Answers
Answer:
Of possible use:
Z
sin ax dx = −
cos ax
a
Z
cos ax dx =
sin ax
a
Z
sin2
ax dx =
x
2
−
sin 2ax
4a
Z
cos2
ax dx =
x
2
+
sin 2ax
4a
Z
x sin ax dx =
sin ax
a
2
−
x cos ax
a
Z
x cos ax dx =
cos ax
a
2
+
x sin ax
a
Z
x
2
sin ax dx =
2x
a
2
sin ax +
2
a
3
−
x
2
a
cos ax Z
x
2
cos ax dx =
2x
a
2
sin ax +
2
a
3
−
x
2
a
cos ax
Z
x
2
cos2
bx dx =
4b
3x
3 + 3(2b
2x
2 − 1) sin 2bx + 6bx cos 2bx
24b
3
Z ∞
0
e
−ax2
dx =
1
2
r
π
a
x
3 − y
3 = (x − y)(x
2 + xy + y
2
)
c = 2.99792458 × 108 m/s
e = 1.602176462 × 10−19 Coul
h = 6.626068 × 10−34 J · s = 4.1356668 × 10−15 eV · s
~ = 1.05457148 × 10−34 J · s = 6.58211814 × 10−16 eV · s
me = 9.10938188 × 10−31 kg = 0.510998903 MeV/c2
mp = 1.67262158 × 10−27 kg = 938.271996 MeV/c2
mn = 1.6749286 × 10−27 kg = 939.565630 Me
Explanation: