PHYSICS

JAMB 2000 - Question 44

Physics 2000 JAMB Past Questions - Question 44: if the frequency of an emitted x-rays is 1.6 x 10¹ hertz accelerating potential is

if the frequency of an emitted x-rays is 1.6 x 10¹ hertz accelerating potential is
A:
B:
C:
D:
Examkits App

Examkit's JAMB CBT App

Practice JAMB offline with our Online, PC and Mobile App

  • ✅ 25+ years of past questions (2000 to 2025)
  • ✅ Video solutions and explanation to questions
  • ✅ E-library
  • ✅ Study by topic
  • ✅ And more.

Correct Answer

B

Explanation

The energy of an emitted X-ray photon can be calculated using the formula:

E = h*fTo calculate the accelerating potential (V) for X-rays, we can use the equation:

E = eV

where E is the energy of the X-rays and e is the elementary charge.

The energy of a photon can be calculated using the equation:

E = hf

where h is Planck's constant (approximately 6.626 x 10^-34 J·s) and f is the frequency of the X-rays.

In this case, the frequency of the emitted X-rays is given as 1.6 x 10^16 Hz. Substituting this value into the equation, we have:

E = (6.626 x 10^-34 J·s) * (1.6 x 10^16 Hz)

Calculating this expression, we find:

E ≈ 1.06 x 10^-17 J

Now, we can calculate the accelerating potential (V) by rearranging the equation:

V = E / e

The elementary charge (e) is approximately 1.602 x 10^-19 C. Substituting this value, we have:

V = (1.06 x 10^-17 J) / (1.602 x 10^-19 C)

Calculating this expression, we find:

V ≈ 66.2 V

Therefore, the accelerating potential for the emitted X-rays is approximately 66.2 volts.

Where:
E is the energy of the photon (in joules)
h is Planck's constant, which is approximately 6.626 x 10^-34 joule seconds
f is the frequency of the X-ray photon (in hertz)

In your case, the frequency (f) is given as 1.6 x 10^16 hertz.

E = (6.626 x 10^-34 J·s) * (1.6 x 10^16 Hz)

E ≈ 1.06 x 10^-17 joules

Now, let's calculate the accelerating potential (V) required to produce this energy for an X-ray photon:

E = e*V

Where:
E is the energy of the photon (in joules)
e is the elementary charge, which is approximately 1.602 x 10^-19 coulombs
V is the accelerating potential (in volts)

1.06 x 10^-17 J = (1.602 x 10^-19 C) * V

Now, solve for V:

V ≈ (1.06 x 10^-17 J) / (1.602 x 10^-19 C)

V ≈ 66.25 x 10^1 V

V ≈ 6625 volts

So, an accelerating potential of approximately 6625 volts is required to produce X-rays with a frequency of 1.6 x 10^16 hertz.