a) An electron in a hydrogen atom has energy E = -3.40 eV, where the zero of ene
ID: 1567925 • Letter: A
Question
a) An electron in a hydrogen atom has energy E = -3.40 eV, where the zero of energy is at the ionization threshold. In the Bohr model, what is the angular momentum of the electron? Express your result as a multiple of h. b) What is the deBroglie wavelength of the electron when it is in this state? c) When the electron is in this state, what is the ratio of the circumference of the orbit of the electron to the deBroglie wavelength of the electron? d) The electron makes a transition from the state with energy E = -3.40 eV to the ground state, that has energy -13.6 eV. What is the wavelength of the photon emitted during this transition?Explanation / Answer
E = -3.4 eV
E = -13.6eV/n^2
-13.6/n^2 = -3.4
n = 2
frm Bohr second postulate
L = mvr = n(h/2pi)
L = 2*(h/2pi)
b) vn = 2.2*10^6/n
n = 2
v = 2.2*10^6/2 = 1.1*10^6 m/s
mass of electron m = 9.1*10^-31 kg
lamda = h/mv
lamda = 6.63*10^-34/(9.1*10^-31*1.1*10^6)
lamda = 6.62*10^-10 m
(c) rn = 0.53*10^-10 n^2
C/lamda = 2pi*r/lamda
= 2*3.14*0.53*10^-10*2^2/(6.62*10^-10)
c/lamda = 2
(d) E1 = -3.4 eV , E2 = -13.6 eV
DE = hc/lamda = E1 - E2
DE = -3.4 +13.6 = 10.2 eV
E(eV) = 1243/lamda(nm)
10.2 = 1243/lamda
lamda = 121.9 nm
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