JEE Main 2014 · 12 Apr · Shift 1 · Chemistry Q31

← JEE Main 2014 · 12 Apr · Shift 1

ChemistrySingle correct+4 / -1Question 31 of 90

If m and e are the mass and charge of the revolving electron in the orbit of radius r for hydrogen atom, the total energy of the revolving electron will be :

\dfrac12\dfrac{e^2}r

-\dfrac{e^2}r

\dfrac{me^2}r

-\dfrac12\dfrac{e^2}r

Answer and worked solution

Official answer: D

In Gaussian units the Coulomb attraction supplies the centripetal force, $\dfrac{mv^2}r=\dfrac{e^2}{r^2}$ , so $K=\dfrac{e^2}{2r}$ and $U=-\dfrac{e^2}r$ .
$E=K+U=-\frac12\frac{e^2}r.$

Topics:Bohr model

← Previous question Next question →

JEE Main 2014 · 12 Apr · Shift 1, Chemistry question 31