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Grade: 12
        if the total energy of an electron in a hydrogen like atom in excited state is -3.4 eV then the de-broglie wavelength of the electron is
6 years ago

Answers : (2)

anonymous
20 Points
							total energy in ground state (t.e) = -13.6ev
hence, n^2 = 13.6/3.4
n=2
v(velocity)= 2pkze^2/nh
hence,  v= 2×3.14×9×10^9×1×(1.6× 10^-19)^2 ÷ 2×6.626 ×10^-34
hence,v=10.96×10^5
p(linear momentum)= mv
de broglie wavelength(wl) =h/p
wl= 6.626×10^-34/9.1×10^-31×10.96×10^5
wl=0.06643× 10^-8
wl= 6.643angstorm
						
6 years ago
Nirmal Singh.
askIITians Faculty
44 Points
							Total Energy K = 3.4 eV = 3.4x1.9x10^-19 joule.
So p = Sqrt (2mK) = sqrt (2*9.1*10^-31*3.4x1.9x10^-19) = 10/84x10^-35 kgm/s
So debroglie wavelength = h/p = 6.63x10^-34 / 10.84x10^-25 = 0.611x10^-9 = 0.611 nm
Regards,
Nirmal Singh
Askiitians faculty
6 years ago
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