Question

Based on equation E = -2.178 x 10^-18 J\((\frac{Z^2}{n^2})\) certain conclusions are written. Which of them is not correct ?

• A. The 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
• B. Larger the value of n, the larger is the orbit radius
• C. Equation can be used to calculate the change in energy when the electron changes orbit
• D. For 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

Answer 1

The Correct Option is D

To solve this question, we need to analyze the given equation and statements based on it. The equation provided is:

E = -2.178 \times 10^{-18} \,J \left(\frac{Z^2}{n^2}\right)

where E is the energy of an electron in a hydrogen-like atom, Z is the atomic number (for hydrogen, Z = 1), and n is the principal quantum number.

Now, let's analyze each option:

1. The 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: This statement is correct. The negative sign indicates that the electron is in a bound state with the nucleus, having less energy than a free electron at infinity.
2. Larger the value of n, the larger is the orbit radius: This statement is correct. As n increases, the principal quantum number dictates that the electron is in a higher energy level, which corresponds to a larger orbit radius.
3. Equation can be used to calculate the change in energy when the electron changes orbit: This is a correct statement. The equation can indeed be used to compute the energy difference when an electron transitions between two energy levels.
4. For 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: This statement is incorrect. Although for n = 1 the energy is more negative than for n = 6, this means the electron is more tightly bound, not loosely. In a hydrogen atom, lower energy means the electron is more tightly bound to the nucleus.

Hence, the incorrect statement is: "For 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". This statement misinterprets the implications of negative energy values.