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If an orbital electron of the hydrogen a...

If an orbital electron of the hydrogen atom jumps from the groud state to a higher energy state, its orbital value where its velcoity is reduced to half its initial value.. If the radius of the electron orbit in the ground state is `r`, then the radius of the new orbit would be:

A

2r

B

4r

C

8r

D

16r

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The correct Answer is:
To solve the problem, we need to determine the new radius of the electron's orbit in a hydrogen atom after it jumps to a higher energy state where its velocity is reduced to half of its initial value. ### Step-by-Step Solution: 1. **Understand the Initial Conditions**: - The electron is initially in the ground state of the hydrogen atom. Let's denote the radius of the electron's orbit in the ground state as \( r \). - The initial velocity of the electron in the ground state is denoted as \( V \). 2. **Determine the New Velocity**: - According to the problem, the new velocity \( V' \) after the electron jumps to a higher energy state is half of the initial velocity: \[ V' = \frac{1}{2} V \] 3. **Relate Velocity to Principal Quantum Number (n)**: - The velocity of an electron in a hydrogen atom is inversely proportional to the principal quantum number \( n \) of the orbit: \[ V \propto \frac{1}{n} \] - Therefore, if the new velocity \( V' \) is half of the initial velocity \( V \), we can set up the following relationship: \[ \frac{V'}{V} = \frac{1}{2} \implies \frac{1}{n'} = \frac{1}{2} \cdot \frac{1}{n} \] - This implies: \[ n' = 2n \] - For the ground state, \( n = 1 \), thus: \[ n' = 2 \cdot 1 = 2 \] 4. **Relate Radius to Principal Quantum Number**: - The radius of the electron's orbit is directly proportional to the square of the principal quantum number \( n \): \[ r \propto n^2 \] - Therefore, the radius of the new orbit \( r' \) can be expressed as: \[ r' = k \cdot n'^2 \] - Since \( r = k \cdot n^2 \), we can relate the two radii: \[ r' = k \cdot (2)^2 = k \cdot 4 = 4r \] 5. **Conclusion**: - The radius of the new orbit \( r' \) is four times the radius of the ground state: \[ r' = 4r \] ### Final Answer: The radius of the new orbit would be \( 4r \).
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