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:

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What is the molar mass of sulfur dioxide (SO$_2)$?

Answer 1

Different approach, using the direct energy difference formula instead of writing each level separately.

The energy released or absorbed between two levels $n_1$ and $n_2$ in hydrogen is given by $\Delta E = x\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$, where x is the ground state ionisation energy.
Here the electron goes from $n_1 = 2$ to $n_2 = 3$, so we need $\Delta E = x\left(\dfrac{1}{4} - \dfrac{1}{9}\right)$.
Convert both fractions to a denominator of 36: $\dfrac{1}{4} = \dfrac{9}{36}$ and $\dfrac{1}{9} = \dfrac{4}{36}$.
So $\Delta E = x\left(\dfrac{9}{36} - \dfrac{4}{36}\right) = x \times \dfrac{5}{36} = \dfrac{5x}{36}$.
This confirms option A. The value is positive because energy must be absorbed to move an electron away from the nucleus to a higher orbit.
Options B and C overshoot badly, they would only make sense if the electron were jumping close to the nucleus from a very high orbit, not from n=2 to n=3. Option D ignores the correct fraction subtraction entirely.

Answer 2

Different approach, using the direct energy difference formula instead of writing each level separately.

The energy released or absorbed between two levels $n_1$ and $n_2$ in hydrogen is given by $\Delta E = x\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$, where x is the ground state ionisation energy.
Here the electron goes from $n_1 = 2$ to $n_2 = 3$, so we need $\Delta E = x\left(\dfrac{1}{4} - \dfrac{1}{9}\right)$.
Convert both fractions to a denominator of 36: $\dfrac{1}{4} = \dfrac{9}{36}$ and $\dfrac{1}{9} = \dfrac{4}{36}$.
So $\Delta E = x\left(\dfrac{9}{36} - \dfrac{4}{36}\right) = x \times \dfrac{5}{36} = \dfrac{5x}{36}$.
This confirms option A. The value is positive because energy must be absorbed to move an electron away from the nucleus to a higher orbit.
Options B and C overshoot badly, they would only make sense if the electron were jumping close to the nucleus from a very high orbit, not from n=2 to n=3. Option D ignores the correct fraction subtraction entirely.

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