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)

As we learnt in

De - Broglie wavelength -

\lambda = \frac{h}{p}= \frac{h}{mv}= \frac{h}{\sqrt{2mE}}

- wherein

h= plank's\: constant

m= mass \: of\: particle

v= speed \: of\: the \: particle

E= Kinetic \: energy \: of \: particle

 

 eV=\frac{hc}{\lambda_{min}}

V=\frac{hc}{e\lambda_{min}}

Take log both side

log\ V=log\frac{hc}{e}-log{\lambda_{min}}

log{\lambda_{min}}=log\frac{hc}{e}-log\ V                            (1)

The correct graph is (1)


Option 1)

This is the correct option.

Option 2)

This is an incorrect option.

Option 3)

This is an incorrect option.

Option 4)

This is an incorrect option.

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