20)Bohr hydrogen atom 21)Wave length Of the following transitions in the Bohr hy

Question

20)Bohr hydrogen atom 21)Wave length Of the following transitions in the Bohr hydrogen atom, the ____________ transition results in the emission of the highest-energy photon. n = 6 rightarrow n = 4 n = 2 rightarrow n = 7 n = 4 rightarrow n = 6 n = 1 rightarrow n = 4 All transitions emit photons of equivalent energy. An electron makes a transition from a Bohr hydrogen orbit of n = 4 to n = 2. The wavelength of light emitted by an hydrogen atom during this transition is ___________ nm. ________________

Explanation / Answer

the equation is directly proportional to:

E = (1/nf^2 1/ni^2)

so calculate:

a)

1/6^2 - 1/4^2 = -0.0347

b)

1/7^2 - 1/2^2 = -0.2295

c)

1/6^2 - 1/4^2 = -0.0347

d)

1/4^2 - 1/1^2 = 0.9375

longest transition is from 1 to 4.. but this is not emision, but absorption

so..

the only option is a) from 6 to 4... since all other are incresa in energy

Q21

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)*(1/4^2 – 1/2 ^2)

E =4.083^-19

For the wavelength:

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)/(4.083*10^-19) = 4.8684*10^-7

WL in nm = (4.8684*10^-7)(10^9) = 486.84 nm

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