Consider a large number of hydrogen atoms, with electrons all initially in the I

Question

Consider a large number of hydrogen atoms, with electrons all initially in the

In a Rutherford scattering experiment, an ?-particle (charge = +2e) heads directly toward a mercury nucleus (charge = +80 e). The ?-particle had a kinetic energy of 5.0 MeV when very far (r rightarrow ?) from the nucleus. Assuming the mercury nucleus to be fixed in space, determine the distance of closest approach. (Hint: Use conservation of energy with PE = keq1q2/r.) fm Consider a large number of hydrogen atoms, with electrons all initially in the n = 4 state. How many different wavelengths would be observed in the emission spectrum of these atoms? There will be different wavelengths observed. What is the longest wavelength that could be observed? nm

Explanation / Answer

1)

K.E. = 5 MeV = = P.E. = keq1q2/r

r = keq1q2/K.E. = 8.99x10^9 * 80 * 2 * 1.6 * 10^(-19) /5*10^(6) = 46.0288 * 10^(-15) m = 46.0288 fm

2)

No. of wavelength = n*(n-1)/2 = 4*(4-1)/2 = 6

3)

1/lambda = Rh[(1/n1^(2)) - (1/n2^(2))]

Rh = 1.09678 x 10-2 nm-1

For n1 = 3

n2 = 4

1/lambda = 1.09678 x 10-2 * [1/9 - 1/16]

= 1.09678 x 10-2 * [0.04861]

= 0.05331 * 10-2 nm-1

longest wavelength = lambda = 18.7566 * 10^(-7) m = 1875.66 nm

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