JEE Main Solved Paper 2017 Question 8
Question: 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 $ {\lambda _{\min }} $ is the smallest possible wavelength of X-ray in the spectrum, the variation of $ \log {\lambda _{\min }} $ with log V is correctly represented in: [JEE Main Solved Paper-2017]
Options:
A) 
B) 
C) 
D) 
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \frac{hc}{{\lambda _{\min }}}=eV $ $ \frac{1}{{\lambda _{\min }}}=\frac{eV}{hc} $ $ \ell n( \frac{1}{{\lambda _{\min }}} )=\ell nV+\ell n\frac{e}{hc} $ $ -n({\lambda _{\min }})=\ell nV+\ell n\frac{e}{hc} $ $ \ell n({\lambda _{\min }})=-\ell nV-\ell n( \frac{e}{hc} ) $ It is a straight line with -ve slope.
