Question

Calculate the energy(J) change associated with an electron transition from n=2 to n=5 in a Bohr hydrogen atom. Put answer in scientific notation.

Answer 1

To find the energy change for an electron moving from n=2 to n=5 in a Bohr hydrogen atom, calculate the energy at both levels and subtract them, yielding an energy change of 4.574 x 10^-19 joules.

Step-by-step explanation:

To calculate the energy change associated with an electron transition from n=2 to n=5 in a Bohr hydrogen atom, we first need to use the formula for the energy levels of hydrogen. This formula is given by En = -13.6 eV / n^2, where En is the energy of the nth level. You can also use the identical equation with energy in joules (J) which is En = -2.178 x 10^-18 J / n^2.

To find the energy change (ΔE) for an electron moving from n=2 to n=5, we calculate the energy of the two levels and then find the difference:

• E2 = -2.178 x 10^-18 J / (2^2) = -2.178 x 10^-18 J / 4 = -0.5445 x 10^-18 J
• E5 = -2.178 x 10^-18 J / (5^2) = -2.178 x 10^-18 J / 25 = -0.08712 x 10^-18 J

Then, calculate the energy change (ΔE):

ΔE = E5 - E2 = (-0.08712 x 10^-18 J) - (-0.5445 x 10^-18 J) = 0.4574 x 10^-18 J = 4.574 x 10^-19 J

The energy change associated with this electron transition is therefore 4.574 x 10^-19 joules.

Answer 2

4.58*10^(-19) J

Step-by-step explanation:

Using the Rydberg's equation:

`(1)/(\lambda) = R * ((1)/(n^2_(final)) - (1)/(n^2_(initial)))`

where

`\lambda` is the wavelength of the photon;

R is the Rydberg's constant = 1.0974*10^7 m^(-1)

final level is 5 and initial level is 2.

`(1)/(\lambda) = 1.0974 * 10^7 * ((1)/(5^2) - (1)/(2^2))`

`(1)/(\lambda) = -2304540 m^(-1)`

`\lambda = -4.339 * 10^(-7) \;m`

Energy change is calculated with the next formula:

E = h*c/λ

where h is the Planck's constant = 6.626*10^(-34) J*s, and c is the speed of light = 299,792,458 m/s

E = 6.626*10^(-34)*299,792,458/-4.339*10^(-7)

E = 4.58*10^(-19) J