Q51 / 180 · NEET 2013 (All India) · Chemistry
Based on equation $E=-2.178 \times 10^{-18} \mathrm{~J}\left(\frac{Z^2}{n^2}\right)$ certain conclusions are written. Which of them is not correct?
- AThe negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus
- BLarger the value of $n$, the larger is the orbit radius
- CEquation can be used to calculate the change in energy when the electron changes orbit
- Correct answerDFor $n=1$ the electron has a more negative energy than it does for $n=6$ which means that the electron is more loosely bound in the smallest allowed orbit
Solution
If $n=1$,
$$
E_1=-2.178 \times 10^{-18} Z^2 \mathrm{~J}
$$
If $n=6$
$$
\begin{aligned}
E_6 & =\frac{-2.178 \times 10^{-18} Z^2}{36} \mathrm{~J} \\
& =6.05 \times 10^{-20} Z^2 \mathrm{~J}
\end{aligned}
$$
From the above calculation, it is abvious that electron has a more negative energy than it does for $n=6$. It means that electron is more strongly bound in the smallest allowed orbit.
Hence, the correct answer is option: 'D'.