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  • An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays.  It produces continuous as well as characteristic X-rays. If λmin is the smallest

An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays.  It produces continuous as well as characteristic X-rays. If λmin is the smallest possible wavelength of X-ray in the spectrum, the variation of log λmin with log V is correctly represented in :

Option: 1


Option: 2


Option: 3


Option: 4


Answers (1)

best_answer

The smallest possible wavelength  \lambda _{min} of an X-ray produced by an electron beam hitting a metallic target is determined by the energy of the electron beam, which is related to the potential difference V used to accelerate the electrons.

As the potential difference V is increased, the minimum wavelength  \lambda _{min}  decreases logarithmically.
This relationship can be expressed as

log \lambda _{min}= -\log V+constant,  

where the constant includes the terms \log \left ( \frac{hc}{e} \right ).

The slope of this relationship is -1, indicating a straight line.

Therefore, the variation of log \lambda_{min} with log V is a straight line with a negative slope of -1.

This relationship is based on the physics of X-ray production, where the energy of an X-ray photon is related to the energy of the electron beam. The minimum wavelength λmin is determined by the minimum energy required to produce an X-ray photon. This minimum energy is related to the potential difference V used to accelerate the electrons.

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himanshu.meshram

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