Question

Consider a silver target in the Coolidge tube to produce x-rays. The accelerating potential is 31kV \[{{E}_{k}}=25.51\]KeV, \[{{E}_{l}}=3.51\] KeV. \[{{\lambda }_{ka}}-{{\lambda }_{\min }}\](in pm). Round off to the nearest integer. (Take hc = 1240 eV nm)

Answer 1

Hint: In this question we have been asked to calculate the difference between wavelengths of X-rays under given condition. To solve this question, we shall first calculate the given wavelengths. We shall use Planck’s equation, which states the relation between energy, frequency and wavelength. Therefore, we shall calculate the wavelengths using the Planck’s equation and get their difference as our final answer.

Formula used:
 \[E=\dfrac{hc}{\lambda }\]

Complete answer:
It is given to us that the accelerating potential is 31 kV. We know that one eV is the energy (kinetic) given to a fundamental charge that is accelerated through a potential of 1 V.
Therefore, we can say that,
\[K.E=31KeV=31\times {{10}^{3}}eV\]
Now using Planck’s equation,
\[E=\dfrac{hc}{\lambda }\]
Therefore, \[\lambda \] can be given by,
\[\lambda =\dfrac{hc}{E}\]
We know \[{{\lambda }_{\min }}\] can be given by
\[{{\lambda }_{\min }}=\dfrac{hc}{K.E}\]
It is given that hc = 1240 eV
Therefore,
\[{{\lambda }_{\min }}=\dfrac{1240}{31\times {{10}^{3}}}\]
Therefore,
\[{{\lambda }_{\min }}=0.04nm\]
Now, we know for \[{{\lambda }_{ka}}\],
\[{{\lambda }_{ka}}=\dfrac{hc}{{{E}_{k}}-{{E}_{l}}}\]
After substituting given values
 We get,
\[{{\lambda }_{ka}}=\dfrac{1240}{(25.51-3.51)\times {{10}^{3}}}\]
On solving,
\[{{\lambda }_{ka}}=\dfrac{0.62}{11}\]
Therefore,
\[{{\lambda }_{ka}}=0.563nm\]
Now taking the difference of (1) and (2)
We get,
\[{{\lambda }_{ka}}-{{\lambda }_{\min }}=0.563-0.04\]
Therefore,
\[{{\lambda }_{ka}}-{{\lambda }_{\min }}=0.523nm\]
Therefore, the correct answer is 0.523 nm.

Note:
A Coolidge tube, also known as X-ray tube is a vacuum tube that converts electrical input power into X-rays. In this tube the cathode is made of spiral filament of incandescent tungsten and the target will serve as the anode. In this tube a cathode will determine the intensity of X-rays and the wavelength will depend on applied voltage.