Electromagnetic Waves

JEE Main level, single correct, medium

165 of 165

Question 165 of 165

An electron beam is accelerated by a potential difference VV to hit a metallic target to produce X−X-rays. It produces continuous as well as characteristic X−X-rays. If λmin{\lambda }_{min} is the smallest possible wavelength of X−X-ray in the spectrum, the variation of log⁡(λmin)\log \left({\lambda }_{min}\right) with log⁡(V)\log \left(V\right) is correctly represented in :

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Answer: (B)

K.E.=eVK.E.=eV,
KE=KE=The kinetic energy of electron

The maximum energy of photon
=K.E.=K.E.

eV=hCλmineV=\dfrac{hC}{{\lambda }_{min}}
⇒λminV=hCe=constant\Rightarrow {\lambda }_{min}V=\dfrac{hC}{e}=constant

Taken
log⁡\log in both sides

ln⁡(λmin⁡)+ln⁡(V)=ln⁡(constant)\ln \left({\lambda }_{\min }\right)+\ln \left(V\right)=\ln \left(constant\right)

⇒ln⁡(λmin⁡)=−ln⁡(V)+ln⁡(constant)\Rightarrow \ln \left({\lambda }_{\min }\right)=-\ln \left(V\right)+\ln (constant)

y=−ax+cy=-ax+c the straight line of negative slope.

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