Use the following information to answer the next two questions. The first four e

Question

Use the following information to answer the next two questions. The first four energy levels for an atom are as follows: Es = -0.54 eV E_3 = -0.95 eV E_2 = -2.14 eV E_1 = -8.57 eV An electron with a kinetic energy of 7.74 eV collides with an atom in its ground state (E_I). After the collision, the atom is left in an excited state and the electron is scattered. Determine the speed of the scattered electron. 2.39 times 10^6 m/s 5.40 times 10^5 m/s 3.19 times 10^5 m/s 2.05 times 10^5 m/s An electron drops from the third energy level (E_3) to the second energy level (E_2) within the atom as shown above. The frequency of the photon associated with this orbital change can be written as a b times 10^td Hz. The values of a, b, c, and d are and

Explanation / Answer

total kinetic energy of electron after collision = 8.57-7.74 = 0.83 eV = 0.83*1.6*10^-19 J

Kinetic energy = (1/2)*m*v^2

m = mass of the electron = 9.11*10^-31 kg

0.83*1.6*10^-19 = (1/2)*9.11*10^-31*v^2

v = 5.4*10^5 m/s

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3)

energy of the released photon = E3 - E1 = -0.95-(-8.57) = 7.62 eV = 7.62*1.6*10^-19 J

energy of photon = hf

h = plancks constant

7.62*1.6*10^-19 = 6.626*10^-34*f

frequency f = 1.8*10^15 Hz

a = 1

b = 8

c = 1

d = 5

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