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NEET CHEMISTRYMedium

From the given equation, certain conclusions are drawn: E=−2.178×10−18 J (Z2n2)E = -2.178 \times 10^{-18} \text{ J } \left(\frac{Z^2}{n^2}\right). Choose the incorrect statement:

A

The negative sign in the equation implies that the energy of an electron bound to the nucleus is lower than it would be if the electron were at an infinite distance from the nucleus.

B

The larger the value of n, the greater the orbit radius.

C

The equation can be utilized to calculate the change in energy when the electron transitions between orbits.

D

For n=1, the electron has more negative energy than it does for n=6. This means that the electron is more loosely bound in the smallest allowed orbit.

Step-by-Step Solution

We analyze the statements based on Bohr's theory:

  1. Significance of Negative Energy: The negative sign indicates that the electron is bound to the nucleus. An electron at infinite distance (n=∞n=\infty) has zero energy. A bound electron has lower (negative) energy. (Statement 1 is Correct).
  2. Radius Relationship: The radius of a Bohr orbit is given by rn∝n2r_n \propto n^2. As nn increases, the radius increases. (Statement 2 is Correct).
  3. Energy Transitions: The energy difference between two levels is ΔE=Efinal−Einitial\Delta E = E_{final} - E_{initial}. This equation allows the calculation of photon energy during transitions. (Statement 3 is Correct).
  4. Binding Energy: For n=1n=1, the energy is the most negative (lowest possible potential well). This corresponds to the most stable and tightly bound state. As nn increases (e.g., n=6n=6), the energy becomes less negative (closer to zero), making the electron less tightly bound. Therefore, saying the electron is "loosely bound" in the smallest orbit (n=1n=1) is false. (Statement 4 is Incorrect).
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