2 Marks Questions

Question 12 Marks

Find the cutoff wavelength for the continuous X-rays coming from an X-ray tube operating at 30kV.

Answer

V = 30KV
$\lambda=\frac{\text{hc}}{\text{E}}=\frac{\text{hc}}{\text{eV}}=\frac{1242\text{eV}-\text{nm}}{\text{e}\times30\times10^{3}}$
$=414\times10^{-4}\text{nm}=41.4\text{pm}$

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Question 22 Marks

When 40kV is applied across an X-ray tube, X-ray is obtained with a maximum frequency of $9.7 \times 10^{18}Hz$. Calculate the value of Planck constant from these data.

Answer

$\lambda=\frac{\text{hc}}{\text{eV}}$
$\text{V}=40\text{Kv}$
$\text{f}=9.7\times10^{18}\text{Hz}$
$\frac{\text{h}}{\text{c}}=\frac{\text{h}}{\text{eV}}$
$\frac{\text{i}}{\text{f}}=\frac{\text{h}}{\text{eV}}$
$\text{h}=\frac{\text{eV}}{\text{f}}\text{V-s}$
$=\frac{\text{eV}}{\text{f}}\text{V-s}=\frac{40\times10^3}{9.7\times10^{18}}$
$=4.12\times10^{-15}\text{eV-s}$

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Question 32 Marks

Find the maximum potential difference which may be applied across an X-ray tube with tungsten target without emitting any characteristic K or L X-ray. The energy levels of the tungsten atom with an electron knocked out are as follows.

Cell containing vacancy	K	L	M
Energy in keV	69.5	11.3	2.3

Answer

Let for, k series emission the potential required = v
$\therefore$ Energy of electrons = ev
This amount of energy ev = energy of L shell
The maximum potential difference that can be applied without emitting any electron is 11.3ev.

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Question 42 Marks

Heat at the rate of 200W is produced in an X-ray tube operating at 20kV. Find the current in the circuit. Assume that only a small fraction of the kinetic energy of electrons is converted into X-rays.

Answer

Heat produced/sec = 200w$\Rightarrow\frac{\text{neV}}{\text{t}}=200\Rightarrow\frac{\text{ne}}{\text{t}}\text{V}=200$
$\Rightarrow\text{i}=200/\text{V}=10\text{mA}$

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Question 52 Marks

X-ray and visible light travel at the same speed in vacuum. Do they travel at the same speed in glass?

Answer

Speed of light in any material medium is inversely proportional to the refractive index of the medium. Since refractive index of glass for X-ray is less than that for visible light, an X-ray will travel at a faster speed than visible light in glass.

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Question 62 Marks

The wavelength of $\text{K}_\alpha$ X-ray of tungsten is 21.3pm. It takes 11.3keV to knock out an electron from the L shell of a tungsten atom. What should be the minimum accelerating voltage across an X-ray tube having tungsten target which allows production of $\text{K}_\alpha$ X-ray?

Answer

$\text{K}_\lambda=21.3\times10^{-12}\text{pm}$ Now $\text{E}_\text{K}-\text{E}_\text{L}=\frac{1242}{21.3\times10^{-3}}=58.309\text{Kev}$ $\text{E}_\text{L}=11.3\text{Kev}$ $\text{E}_\text{K}=58.309=11.3=69.609\text{Kev}$ Now, Ve = 69.609KeVor V = 69.609KV.

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Question 72 Marks

Why is exposure to X-rays injurious to health but not exposure to visible light, when both are electromagnetic waves?

Answer

X-rays have more penetrating power compared to visible light. As a result, they can penetrate the human body and can also damage the cells of the body. Prolonged exposure to X-rays can lead to cancer or genetic defects.

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Question 82 Marks

The distance between the cathode (filament) and the target in an X-ray tube is 1.5m. If the cutoff wavelength is 30pm, find the electric field between the cathode and the target.

Answer

$\text{d}=1.5\text{m},$
$\lambda=30\text{pm}=30\times10^{-3}\text{nm}$
$\text{E}=\frac{\text{hc}}{\lambda}=\frac{1242}{30\times10^{-3}}$
$=41.4\times10^3\text{eV}$
Electric field $=\frac{\text{V}}{\text{d}}=\frac{41.4\times10^3}{1.5}$
$=27.6\times10^3\text{V/m}=27.6\text{KV/m}$

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Question 92 Marks

The $\text{K}_\beta$ X-ray of argon has a wavelength of 0.36nm. The minimum energy needed to ionize an argon atom is 16eV. Find the energy needed to knock out an electron from the K shell of an argon atom.

Answer

$\lambda=0.36\text{nm}$
$\text{E}=\frac{1242}{0.36}=3450\text{eV}(\text{E}_\text{M}-\text{E}_\text{K})$
Energy needed to ionize an organ atom = 16eV
Energy needed to knock out an electron from K-shell
= (3450 + 16)eV = 3466eV = 3.466KeV.

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