Problem:

An excited hydrogen atom could, in principle, have a radius of 2.50 mm .
________
Part A:

What would be the value of n for a Bohr orbit of this size?

Answer--> n= 6.87×103
_______
Part B

What would its energy be?

Answer--> E=_______eV

[Note: I got 4.61x10^-26 for part B but I got it wrong 3 times. Please help]

Part A answer is n= 6.87x10^3

To solve this problem, we can use the formula for the radius of the Bohr orbit and the formula for the energy of a hydrogen atom in the nth orbit.

The formula for the radius of the Bohr orbit is:

r = (0.529 × n^2) / Z

where r is the radius of the Bohr orbit, n is the principal quantum number, and Z is the atomic number of the nucleus (which is 1 for hydrogen).

Part A:
We are given the radius of the Bohr orbit as 2.50 mm, so we can plug this value into the formula and solve for n.

2.50 mm = (0.529 × n^2) / 1
2.50 mm × 1 = 0.529 × n^2
n^2 = (2.50 mm × 1) / 0.529
n^2 = 4.7294
n = sqrt(4.7294)
n ≈ 2.18

Therefore, n ≈ 2.18.

Part B:
To calculate the energy of the excited hydrogen atom, we can use the formula for the energy:

E = (-13.6 eV) / n^2

We can substitute the value of n that we found in Part A into this equation and solve for E.

E = (-13.6 eV) / (2.18)^2
E = (-13.6 eV) / 4.7524
E ≈ -13.6 / 4.7524 eV
E ≈ -2.863 eV

Therefore, the energy of the excited hydrogen atom is approximately -2.863 eV.

To solve this problem, we need to use the formula for the radius of the Bohr orbit and the formula for the energy of an excited hydrogen atom.

The formula for the radius of an electron in the nth energy level of the hydrogen atom (also known as the Bohr radius) is given by:

r = (0.529 × 10^-10 m) * n^2 / Z

where r is the radius of the orbit, n is the principal quantum number, and Z is the atomic number (which is 1 for hydrogen).

In Part A, we are given the radius (r) of the excited hydrogen atom as 2.50 mm, which we need to convert to meters:

r = 2.50 mm = 2.50 × 10^-3 m

To find the value of n, we can rearrange the formula for the Bohr orbit radius:

n^2 = (r * Z) / 0.529 × 10^-10 m

n^2 = (2.50 × 10^-3 m * 1) / (0.529 × 10^-10 m)

n^2 = (2.50 × 10^-3 m) / (0.529 × 10^-10 m)

n^2 = 4.725 × 10^6

Taking the square root of both sides gives:

n = √(4.725 × 10^6)
n = 2.17 × 10^3

Therefore, the value of n for a Bohr orbit of this size is approximately 2,170.

In Part B, we need to calculate the energy of the excited hydrogen atom. The formula for the energy (E) of an electron in the nth energy level of the hydrogen atom is given by:

E = -(13.6 eV / n^2)

where E is the energy in electron volts (eV) and n is the principal quantum number.

Plugging in the value of n we found in Part A:

E = -(13.6 eV / (2.17 × 10^3)^2)

E = -(13.6 eV / 4.7089 × 10^6)

E = -2.89 × 10^-9 eV

Therefore, the energy of the excited hydrogen atom is approximately -2.89 × 10^-9 eV.

I'm sorry for the confusion, but the previous answer you provided, 4.61 × 10^-26 eV, seems to be incorrect. The correct answer for Part B is -2.89 × 10^-9 eV.

Rn = n^2 * R1

2.50E-3 m = n^2 * .53E-10 m

En = (13.6 / n^2) eV