An electron in a hydrogen atom makes a transition from an orbital n=4 to an orbi

Question

An electron in a hydrogen atom makes a transition from an orbital n=4 to an orbital n=1. The change in energy that occurs per mole of hydrogen atoms when it undergoes this transition is Blank 1 kJ/mol. (0 dec places). (Hint: the magnitude of the sign matters, i.e. + or -)

The wavelength of light emitted by the transition described above is Blank 2 nm (0 dec places). Hint: The unit of the energy of transition is kJ/mol.

In the Bohr model, each transition corresponds to a spectral line. An electron in a hydrogen atom makes a transition from an orbital n = 4 to an orbital n = 1. The change in energy that occurs per mole of hydrogen atoms when it undergoes this transition is kJ/mol. (0 dec places). The wavelength of light emitted by the transition described above is nm (0 dec places). Because of the high voltage required to get the discharge tube to light up, you should not keep the lamps on for more than 1 minute. In order to see the spectral lines without interference from the overhead lights, you should hold the tube very close to the lamps.

Explanation / Answer

Q3.

n = 4 to n = 1

Apply Rydberg Formula

E = R*(1/nf^2 – 1/ni ^2)

R = -2.178*10^-18 J

Nf = final stage/level

Ni = initial stage/level

E = Energy per unit (i.e. J/photon)

E = (-2.178*10^-18)*(1/1^2 – 1/4 ^2)

E = 2.0418*10^-18 J/particle emited

1 mol = 6.022*10^3 particles

so..

E = 2.0418*10^-18 J/particle emited * 6.022*10^3 particles = 1229571.96J/mol = 1229.5 kJ/mol

for wavelngth

WL = h c / E

h = Planck Constant = 6.626*10^-34 J s

c = speed of particle (i.e. light) = 3*10^8 m/s

E = energy per particle J/photon

WL = (6.626*10^-34)(3*10^8)/( 2.0418*10^-18

WL = 9.7355*10^-8 m

to nanometers:

WL = ( 9.7355*10^-8 )(10^9) = 97.355 nm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.