For He+, a transition takes place from the orbit of radius 105.8 pm to the orbit of radius 26.45 pm.

Q: Why do we use $Z=2$ for Helium?

Helium is the second element in the periodic table, meaning its nucleus contains 2 protons. The $Z$ in Bohr's formulas represents the nuclear charge, which is directly equal to the number of protons.

Q: Why is the Rydberg constant given in Joules instead of $m^{-1}$?

The Rydberg constant can be expressed in different units depending on the formula used. When expressed in Joules ($2.2 \times 10^{-18} \text{ J}$), it represents the ionization energy of hydrogen. The formula $\Delta E = R_H Z^2 (\dots)$ uses this energy value directly.

Q: How did we know which orbit was $n_1$ and which was $n_2$?

The problem states the transition is *from* $105.8 \text{ pm}$ *to* $26.45 \text{ pm}$. The larger radius corresponds to the higher energy level (initial state, $n_2$), and the smaller radius corresponds to the lower energy level (final state, $n_1$).

Q: Can I use the formula $\frac{1}{\lambda} = R Z^2 (\dots)$ directly?

Yes, but you would need the Rydberg constant in terms of inverse meters ($R \approx 1.097 \times 10^7 \text{ m}^{-1}$). Since the problem specifically provided $R_H$ in Joules, it's safer and more precise to use the energy equation $\frac{hc}{\lambda} = \Delta E$.

Q: Why did we convert $3 \times 10^{-8} \text{ m}$ to $30 \times 10^{-9} \text{ m}$?

The question specifically asked for the wavelength in nanometers (nm). One nanometer is $10^{-9}$ meters. By rewriting $3 \times 10^{-8}$ as $30 \times 10^{-9}$, we can directly read off the value 30 as the answer in nm.

Question

For $	ext{He}^+$, a transition takes place from the orbit of radius $105.8	ext{ pm}$ to the orbit of radius $26.45	ext{ pm}$.
The wavelength (in nm) of the emitted photon during the transition is ______.
[Use:
Bohr radius, $a = 52.9	ext{ pm}$
Rydberg constant, $R_H = 2.2 	imes 10^{-18}	ext{ J}$
Planck's constant, $h = 6.6 	imes 10^{-34}	ext{ J s}$
Speed of light, $c = 3 	imes 10^8	ext{ m s}^{-1}$]

Sigma Percentile · Accepted Answer

The quantum world is full of fascinating phenomena, and one of the most beautiful is the emission of light when an electron transitions between energy levels. In this problem, we are tasked with finding the wavelength of a photon emitted by a singly ionized helium atom ($	ext{He}^+$) as its electron jumps from a larger orbit to a smaller one.

Let's break down the physics and the math behind this quantum leap!

### Decoding the Orbits

Before we can calculate the energy of the emitted photon, we need to know exactly *where* the electron is jumping from and where it is landing. The problem gives us the radii of the two orbits: an initial radius of $105.8 	ext{ pm}$ and a final radius of $26.45 	ext{ pm}$.

To translate these physical distances into quantum numbers, we use the Bohr radius formula for hydrogen-like species:
$$r_n = a \frac{n^2}{Z}$$
Here, $a$ is the Bohr radius ($52.9 	ext{ pm}$), $n$ is the principal quantum number (the orbit number), and $Z$ is the atomic number. Since we are dealing with a helium ion ($	ext{He}^+$), the atomic number $Z = 2$.

Let's find the initial orbit, $n_2$. We plug in our known values:
$$105.8 = 52.9 	imes \frac{n_2^2}{2}$$
Notice how beautifully the numbers are set up! $105.8$ is exactly double $52.9$. Dividing both sides by $52.9$ gives us $2$. Multiplying by the $2$ from the denominator yields:
$$n_2^2 = 4 \implies n_2 = 2$$
So, the electron starts its journey in the second orbit.

Now, let's find the final orbit, $n_1$, using the smaller radius:
$$26.45 = 52.9 	imes \frac{n_1^2}{2}$$
This time, $26.45$ is exactly half of $52.9$. Dividing both sides by $52.9$ gives us $0.5$, or $\frac{1}{2}$. 
$$\frac{1}{2} = \frac{n_1^2}{2}$$
The twos cancel out perfectly, leaving us with:
$$n_1^2 = 1 \implies n_1 = 1$$
The electron is dropping from the $n=2$ state down to the ground state, $n=1$.

### The Energy of the Jump

When an electron drops to a lower energy level, it must shed its excess energy. It does this by emitting a photon. The energy of this photon ($\Delta E$) is directly related to its wavelength ($\lambda$) by the Planck-Einstein relation, $\Delta E = \frac{hc}{\lambda}$.

To find this energy difference, we use the Rydberg formula. However, we must pay close attention to the units! The problem provides the Rydberg constant, $R_H$, in Joules ($2.2 	imes 10^{-18} 	ext{ J}$). This means the constant represents an energy value, so our formula looks like this:
$$\frac{hc}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$$

Let's substitute all our known values into this master equation:
$$\frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{\lambda} = 2.2 	imes 10^{-18} 	imes 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} ight)$$

### The Final Calculation

Now comes the algebra. Let's simplify the left side first. Multiplying Planck's constant by the speed of light gives:
$$6.6 	imes 3 = 19.8$$
Combining the powers of ten ($10^{-34} 	imes 10^8$) gives $10^{-26}$. So the left side is $\frac{19.8 	imes 10^{-26}}{\lambda}$.

On the right side, we evaluate the terms inside the parentheses:
$$\left( 1 - \frac{1}{4} ight) = \frac{3}{4}$$
Our equation now looks like this:
$$\frac{19.8 	imes 10^{-26}}{\lambda} = 2.2 	imes 10^{-18} 	imes 4 	imes \frac{3}{4}$$
The $4$ from $Z^2$ and the $4$ in the denominator cancel out beautifully! We are left with:
$$\frac{19.8 	imes 10^{-26}}{\lambda} = 2.2 	imes 10^{-18} 	imes 3$$
$$\frac{19.8 	imes 10^{-26}}{\lambda} = 6.6 	imes 10^{-18}$$

Finally, we isolate $\lambda$:
$$\lambda = \frac{19.8 	imes 10^{-26}}{6.6 	imes 10^{-18}}$$
Since $19.8$ is exactly three times $6.6$, the division yields $3$. Subtracting the exponents ($-26 - (-18)$) gives $-8$.
$$\lambda = 3 	imes 10^{-8} 	ext{ m}$$

The question specifically asks for the wavelength in nanometers ($	ext{nm}$). Since $1 	ext{ nm} = 10^{-9} 	ext{ m}$, we can rewrite our answer by shifting the decimal point:
$$\lambda = 30 	imes 10^{-9} 	ext{ m} = \mathbf{30 	ext{ nm}}$$

And there we have it! The emitted photon has a wavelength of $30 	ext{ nm}$, placing it in the extreme ultraviolet region of the electromagnetic spectrum.

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Transition}

\text{Bohr Radius Formula}

\text{Finding Initial Orbit } (n_2)

\text{Calculating } n_2

\text{Finding Final Orbit } (n_1)

\text{Calculating } n_1

\text{Rydberg Equation for Energy}

\text{Substituting Values}

\text{Solving for } \lambda

\text{Final Wavelength in nm}

\text{Looking Ahead}

The Sigma Insight: Bohr's Model

Decoding the Orbits

The Energy of the Jump

The Final Calculation

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Which of the following is the energy of a possible excited state of hydrogen?

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Visualizing the Transition} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Bohr Radius Formula} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Finding Initial Orbit } (n_2) .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Calculating } n_2 .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Finding Final Orbit } (n_1) .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Calculating } n_1 .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Rydberg Equation for Energy} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Substituting Values} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Solving for } \lambda .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Final Wavelength in nm} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Looking Ahead} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }