A) \[s=\frac{\frac{1}{2}\times m\times {{(10)}^{2}}}{\mu mg}\]
B) \[=\frac{50}{0.5\times 10}=10m.\]
C) \[F=\mu R\]
D) \[R=w=250N.\]
Correct Answer: C
Solution :
Key Idea: Atomic number of helium is twice that of hydrogen. When e, m are the charge and mass of electron in the nth orbit, h is Plancks constant and Z is the atomic number, then the energy of the electron in the electron in the nth orbit is \[E=-\frac{M\,{{Z}^{2}}\,{{e}^{4}}}{8{{\varepsilon }_{o}}{{h}^{2}}}.\frac{1}{{{n}^{2}}}\] Given, \[{{Z}_{H}}=1,\,\,{{Z}_{He}}=2\] \[\therefore \] \[\frac{{{E}_{H}}}{{{E}_{He}}}=\frac{Z_{H}^{2}}{Z_{He}^{2}}=\frac{1}{4}\] \[\Rightarrow \] \[{{E}_{He}}=4{{E}_{H}}\] Given, \[{{E}_{H}}={{E}_{n}}\] \[{{E}_{He}}=4{{E}_{n}}\]
You need to login to perform this action.
You will be redirected in
3 sec
