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Question: The voltage applied to an x-ray to be is 18 KV. The maximum mass of photon emitted by the X – ray tu...

The voltage applied to an x-ray to be is 18 KV. The maximum mass of photon emitted by the X – ray tube will be:
A.2×10−13 kg2\times {{10}^{-13}}\text{ kg}
B. 3.2×10−32 kg\text{3}\text{.2}\times {{10}^{-32}}\text{ kg}
C. 6.4×10−32 kg\text{6}\text{.4}\times {{10}^{-32}}\text{ kg}
D. 9.1×10−31 kg\text{9}\text{.1}\times {{10}^{-31}}\text{ kg}

Explanation

Solution

Basically, in X-ray the voltage applied between cathode and anode. In this question we equate the energy of Photon with the maximum energy of an electron. The maximum energy of a photon is mc2m{{c}^{2}}.

Complete step by step answer:
X-rays were produced by Wilhelm Roentgen in 1895 while testing whether cathode rays pass through glass. When the voltage is applied through the x-ray tube or between cathode and anode then maximum energy of the photon remains equal to the maximum energy of the photon.

Given applied voltage =18KV=1⋅8×103 volt=18KV=1\cdot 8\times {{10}^{3}}\text{ volt}
As we know the energy of photon=mc2=m{{c}^{2}}
Where, m = mass of photon
c = speed of light
Also the maximum energy of the photon = maximum energy of electron.
mc2=hcλ=evm{{c}^{2}}=\dfrac{hc}{\lambda }=ev
⇒mc2=ev\Rightarrow m{{c}^{2}}=ev
⇒m=evc2\Rightarrow m=\dfrac{ev}{{{c}^{2}}}
e=1.6×10−19ce=1.6\times {{10}^{-19}}c
⇒m=1.6×10−19×1.8×103(3×108)2\Rightarrow m=\dfrac{1.6\times {{10}^{-19}}\times 1.8\times {{10}^{3}}}{{{\left( 3\times {{10}^{8}} \right)}^{2}}}
⇒m=3.2×10−32 kg\Rightarrow m=3.2\times {{10}^{-32}}\text{ kg}
Hence, the maximum mass of photon is 3.2×10−32 kg3.2\times {{10}^{-32}}\text{ kg}

Therefore, the correct choice is (B) 3.2×10−32 kg3.2\times {{10}^{-32}}\text{ kg}

Note:
When a voltage is applied to an X – ray tube (b/w cathode and anode) then the maximum energy of the emitted photon remains equal to the energy of the electron.
mc2=hcλ=eVm{{c}^{2}}=\dfrac{hc}{\lambda }=eV