Chemistry Q121 – AP-EAMCET 2026

Home Chemistry Structure of AtomQuestion #596

Wavelength of a photon emitted during electron transition from

n = 4

state to

n = 2

state in the hydrogen atom is

x

nm. Wavelength of a photon emitted during electron transition from

n = 4

state to

n = 1

state in the same atom is

y

nm.

\frac{y}{x}

is equal to

Clue 1 of 3: Atomic Transitions and Energy

When an electron transitions from a higher energy state to a lower energy state in an atom, a photon is emitted with energy equal to the energy difference between the states.

🧠 Full Solution Path

Step 1: Recall the Rydberg Formula✦ Active

The wavelength $λ$ of a photon emitted during an electron transition in a hydrogen atom is given by the Rydberg formula:

\frac{1}{λ} = R_{H} (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})

where $R_{H}$ is the Rydberg constant, $n_{i}$ is the initial principal quantum number, and $n_{f}$ is the final principal quantum number.

Step 2: Calculate

1 / x

for the first transition○ Expand

For the first transition, the electron goes from $n_{i} = 4$ to $n_{f} = 2$ , and the emitted wavelength is $x$ nm. Substitute these values into the Rydberg formula:

\frac{1}{x} = R_{H} (\frac{1}{2^{2}} - \frac{1}{4^{2}})

Simplify the expression:

\frac{1}{x} = R_{H} (\frac{1}{4} - \frac{1}{16})

\frac{1}{x} = R_{H} (\frac{4 - 1}{16}) = R_{H} (\frac{3}{16}) (1)

💡 Teacher's Secret Hint

Ensure to use $n_{f}$ as the lower energy state and $n_{i}$ as the higher energy state for emitted photons.

Step 3: Calculate

1 / y

for the second transition○ Expand

For the second transition, the electron goes from $n_{i} = 4$ to $n_{f} = 1$ , and the emitted wavelength is $y$ nm. Substitute these values into the Rydberg formula:

\frac{1}{y} = R_{H} (\frac{1}{1^{2}} - \frac{1}{4^{2}})

Simplify the expression:

\frac{1}{y} = R_{H} (\frac{1}{1} - \frac{1}{16})

\frac{1}{y} = R_{H} (\frac{16 - 1}{16}) = R_{H} (\frac{15}{16}) (2)

Step 4: Calculate the ratio

\frac{y}{x}

○ Expand

To find the ratio $\frac{y}{x}$ , we can express $x$ and $y$ from equations (1) and (2) respectively:

x = \frac{16}{3 R_{H}}

y = \frac{16}{15 R_{H}}

Now, compute the ratio $\frac{y}{x}$ :

\frac{y}{x} = \frac{\frac{16}{15 R_{H}}}{\frac{16}{3 R_{H}}}

\frac{y}{x} = \frac{16}{15 R_{H}} \times \frac{3 R_{H}}{16}

\frac{y}{x} = \frac{3}{15}

\frac{y}{x} = \frac{1}{5} = 0.2

💡 Teacher's Secret Hint

Notice how the Rydberg constant $R_{H}$ cancels out in the ratio, which is expected since the question asks for a relative value and $R_{H}$ is a universal constant for hydrogen.

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Chemistry Question 121 – AP-EAMCET 2026

Clue 1 of 3: Atomic Transitions and Energy

🧠 Full Solution Path