Nuclear Physics And Radioactivity Question 42
Question: An electron passing through a potential difference of 4.9 V collides with a mercury atom and transfers it to the first excited state. What is the wavelength of a photon corresponding to the transition of the mercury atom to its normal state [AMU (Med.) 2002]
Options:
A) 2050 Å
B) 2240 Å
C) 2525 Å
D) 2935 Å
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{hc}{\lambda }=E $ = eV
$ \Rightarrow \lambda =\frac{hc}{eV}-\frac{6.6\times {{10}^{-34}}\times 3\times 10^{8}}{1.6\times {{10}^{-19}}\times 4.9}=2525\overset{\circ }{\mathop{A}}, $
BETA
