An electron is excited from a lower energy state to a higher energy state in a hydrogen atom. Which of the following quantity/quantities decreases/decrease in the excitation ?

Options

A. potential energy
B. Angular speed
C. Kinetic energy
D. Angularn momentum

Answer

To determine which quantities decrease when an electron is excited from a lower energy state to a higher energy state in a hydrogen atom, let's analyze the situation step by step.

### Step 1: Understand the Energy States
In a hydrogen atom, electrons occupy discrete energy levels (or states). When an electron is excited, it moves from a lower energy level (n1) to a higher energy level (n2).

### Step 2: Analyze Kinetic Energy
The kinetic energy (KE) of an electron in a hydrogen atom can be expressed in terms of its velocity (v) as:
\[ KE = \frac{1}{2}mv^2 \]
As the electron moves to a higher energy state, the radius (r) of its orbit increases. The centripetal force required to keep the electron in orbit is provided by the electrostatic force between the electron and the proton.

Using the formula:
\[ F_{centripetal} = \frac{mv^2}{r} = \frac{K \cdot Ze^2}{r^2} \]
where K is Coulomb's constant, Z is the atomic number (1 for hydrogen), and e is the charge of the electron.

As the radius increases (when moving to a higher energy state), the velocity of the electron decreases, leading to a decrease in kinetic energy.

### Step 3: Analyze Potential Energy
The potential energy (PE) of the electron in the hydrogen atom is given by:
\[ PE = -\frac{K \cdot Ze^2}{r} \]
As the radius r increases when the electron is excited to a higher energy state, the magnitude of the potential energy decreases (since it becomes less negative), meaning that the potential energy actually increases.

### Step 4: Analyze Angular Momentum
The angular momentum (L) of the electron is quantized and given by:
\[ L = mvr = n\frac{h}{2\pi} \]
where n is the principal quantum number. As the electron moves to a higher energy state, n increases, which means that angular momentum also increases.

### Step 5: Analyze Angular Speed
Angular speed (ω) can be expressed as:
\[ \omega = \frac{v}{r} \]
Since the velocity v decreases and the radius r increases, the angular speed ω will also decrease.

### Conclusion
From the analysis:
- Kinetic Energy (KE) decreases.
- Potential Energy (PE) increases.
- Angular Momentum (L) increases.
- Angular Speed (ω) decreases.

Thus, the quantities that decrease during the excitation of the electron are:
- Kinetic Energy
- Angular Speed

### Final Answer
The quantities that decrease are:
- Kinetic Energy
- Angular Speed