Nuclear Physics And Radioactivity Question 34
Question: In a hypothetical Bohr hydrogen, the mass of the electron is doubled. The energy $ E_{0} $ and the radius $ r_{0} $ of the first orbit will be ( $ a_{0} $ is the Bohr radius) [Roorkee 1992]
Options:
A) $ E_{0}=-\ 27.2\ eV;\ r_{0}=a_{0}/2 $
B) $ E_{0}=-\ 27.2\ eV;\ r_{0}=a_{0} $
C) $ E_{0}=-13.6\ eV;\ r_{0}=a_{0}/2 $
D) $ E_{0}=-13.6\ eV;\ r_{0}=a_{0} $
Show Answer
Answer:
Correct Answer: A
Solution:
Here radius of electron orbit $ r\propto 1/m $ and energy E µ m, where m is the mass of the electron. Hence energy of hypothetical atom $ E_{0}=2\times (-13.6,eV)=-27.2,eV $ and radius $ r_{0}=\frac{a_{0}}{2} $
BETA
