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
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.

