Question

If the ionisation energy of hydrogen in its ground state is x kJ per mole, the energy needed for an electron to jump from n = 2 to n = 3 is:

• A. 5x/36
• B. 5x
• C. 7.2x
• D. x/6

Hint:

Use E_n = minus x by n squared, then subtract E2 from E3.

Answer 1

The Correct Option is A

Step 1: Write the energy formula for hydrogen atom.
For a hydrogen atom, the energy of an electron in orbit n is \(E_n = -\dfrac{x}{n^2}\), where x is the ionisation energy from the ground state (n = 1 to n = infinity).

Step 2: Find energy at n = 2 and n = 3.
\(E_2 = -\dfrac{x}{4}\)
\(E_3 = -\dfrac{x}{9}\)

Step 3: Find the energy needed to jump from n = 2 to n = 3.
This is simply \(E_3 - E_2\), because the electron absorbs energy to move to a higher level.
\(\Delta E = -\dfrac{x}{9} - \left(-\dfrac{x}{4}\right) = \dfrac{x}{4} - \dfrac{x}{9}\)

Step 4: Take the LCM of 4 and 9, which is 36.
\(\dfrac{x}{4} = \dfrac{9x}{36}\), and \(\dfrac{x}{9} = \dfrac{4x}{36}\)
\(\Delta E = \dfrac{9x}{36} - \dfrac{4x}{36} = \dfrac{5x}{36}\)

So option A, 5x/36, is correct.

Why the other options are wrong: Option B (5x) and option C (7.2x) are far too large, they come from mixing up ground state ionisation energy with a small orbit jump. Option D (x/6) does not match the actual subtraction of 1/4 and 1/9. Only careful subtraction with the correct sign gives 5x/36.