the first excited state of the nucleus of uranium-235 is 0.051 MeV above the ground state. what is the wavelength of the photon emitted when the nucleus makes a transition to the ground state? (in meters)

Convert 0.051 MeV to energy in Joules.

1 eV = 1.6*10^-19 J
1 MeV = 1.6*10^-13 J

Call it E.

Then use
(wavelength) = E/(h c)

where h is Planck's constant and c is the speed of light.

drwls made one crucial error.

Wavelength equals (h c) / E, because who in their right mind would think that the wavelength is directly proportional to energy?

To calculate the wavelength of the photon emitted during the transition from the first excited state to the ground state of uranium-235, we can use the energy-wavelength relationship given by the equation:

E = hc/λ

where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J·s or 4.135 x 10^-15 eV·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the photon.

Given that the energy difference between the first excited state and the ground state is 0.051 MeV, we first need to convert it to the corresponding energy in joules:

1 electronvolt (eV) = 1.602 x 10^-19 Joules (J)

0.051 MeV × (1.602 x 10^-19 J/eV) = 8.181 x 10^-21 J

Now, we can substitute the values into the equation to solve for the wavelength:

8.181 x 10^-21 J = (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s) / λ

Simplifying the equation, we can solve for λ:

λ = (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s) / (8.181 x 10^-21 J)

Calculating the value of λ gives:

λ ≈ 24.162 nm

Therefore, the wavelength of the photon emitted during the transition from the first excited state to the ground state of uranium-235 is approximately 24.162 nanometers (nm).

To determine the wavelength of the photon emitted when the uranium-235 nucleus transitions from the first excited state to the ground state, we can use the energy-wavelength relationship derived from the photon's momentum:

E = hc/λ,

where E is the energy of the photon, h is Planck's constant (6.62607015 × 10^-34 J⋅s), c is the speed of light (2.998 × 10^8 m/s), and λ is the wavelength of the photon.

First, we need to convert the energy difference given (0.051 MeV) into joules. We know that 1 MeV is equivalent to 1.602176634 × 10^-13 Joules.

Converting the given energy difference into joules:

0.051 MeV × (1.602176634 × 10^-13 J/1 MeV) = 8.201896234 × 10^-15 J.

Now, we can rearrange the energy-wavelength relationship equation to solve for λ:

λ = hc/E.

Substituting the known values:

λ = (6.62607015 × 10^-34 J⋅s × 2.998 × 10^8 m/s) / (8.201896234 × 10^-15 J).

Calculating λ:

λ = 2.426028084 × 10^-16 m.

Therefore, the wavelength of the photon emitted when the uranium-235 nucleus transitions from the first excited state to the ground state is approximately 2.426 × 10^-16 meters.