Question

The X-Ray tube is operated at $50kV.$ The minimum wavelength produce will be:
(A) $0.5\mathop A\limits^0 $
(B) $0.75\mathop A\limits^0 $
(C) $0.25\mathop A\limits^0 $
(D) $1.0\mathop A\limits^0 $

Answer 1

Hint: In order to solve this question, we will first convert the given potential difference value from Kilo-volts into volts, and then using the general formula of the wavelength of X-ray we will solve for the value of wavelength produced by the X-ray tube in the units of angstrom.

Formula used:
The wavelength produced by the X-ray tube is calculated using the formula
$\lambda = \dfrac{{hc}}{{eV}}$
where,
$h = 6.626 \times {10^{ - 34}}Js$ is known as the Planck’s constant
$c = 3.0 \times {10^8}m{s^{ - 1}}$ is the speed of light in vacuum
$e = 1.6 \times {10^{ - 19}}C$ is the charge on an electron
$V$ is the potential difference on which X-ray tube is operated and $\lambda $ is the wavelength produced by the X-ray tube.

Complete answer:
According to the question, we have given that X-ray tube is operated at a potential difference of $V = 50kV$ which in volts will be $V = 50,000V$, in order to find the wavelength produced by the X-ray tube, we can use the formula as $\lambda = \dfrac{{hc}}{{eV}}$ and on putting the values of all the known parameters and solving for the value of wavelength we get,
$
  \lambda = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.6 \times {{10}^{ - 19}} \times 50000}} \\
  \lambda = 2.48 \times {10^{ - 9}} \\
  \lambda = 0.25\mathop A\limits^0 \\
 $
So, the value of the wavelength produced by the X-ray tube is $\lambda = 0.25\mathop A\limits^0 $

Hence, the correct answer is option (C) $0.25\mathop A\limits^0 $

Note:It should be remembered that the basic unit of conversion from kilo-volts to volts is used as $1kV = 1000V$ and angstrom is the smaller unit of measuring the length and here the conversion relation of angstrom is used as $1\mathop A\limits^0 = {10^{ - 10}}m$, an X-ray tube is a vacuum tube which usually converts the electrical input into X rays.