The first three energy levels of the fictitious element X are shown in Figure P3

Question

The first three energy levels of the fictitious element X are shown in Figure P38.55, in which E1 = -10.4 eV, E2 = -4.8 eV, and E3 = -3.2 eV.

   (i) (ii) (a) ? (nm) 2      nm 3      nm (b) type
of light 4---Select---ultravioletvisibleinfrared      5---Select---ultravioletvisibleinfrared      The first three energy levels of the fictitious element X are shown in Figure P38.55, in which E1 = -10.4 eV, E2 = -4.8 eV, and E3 = -3.2 eV. What is the ionization energy of element X? Calculate the (i) shortest and (ii) next-shortest wavelengths observed in the absorption spectrum of element X. (Enter these into the table below.) Specify whether each of your wavelengths in part b corresponds to ultraviolet, visible, or infrared light. An electron with a speed of 1.7 times 106 m/s collides with an atom of element X. Shortly afterward, the atom emits a 776 nm photon. What was the electron's speed after the collision? Assume that, because the atom is so much more massive than the electron, the recoil of the atom is negligible. (Hint: The energy of the photon is not the energy transferred to the atom in the collision.)

Explanation / Answer

1) ionization energy = Einiinity - E1

I.E =0 +10.4

I.E = 10.4 eV

so the ionization energy is 10.4 eV

2) E = hc / lamda

E = E2 -E1 = -4.8 + 10.4 = 5.6 eV

lamda 1 = 6.626 x 10-34 x 3 x 108 / 5.6 x 1.6 x 10-19

lamda 1 = 2.21 x 10-7 m

lamda 1 = 221.85 nm

lamda 2 = 6.626 x 10-34 x 3 x 108 / 7.2 x 1.6 x 10-19

lamda 2 = 172.55 nm

shortest is 172.55 nm , ultraviolet

next shortest is 221.85 nm , ultraviolet

3) by the law of conservation of energy

1/2 m V2 + hc/lamda = 1/2 mv2

6.626 x 10-34 x 3 x 108 / 776 x 10-9 = 1/2 m ( v2- V2)

2.56 x 10-19 = m ( v2-V2) /2

v2-V2 = 2.56 x 10-19 x 2 / 9.1 x 10-31

v2 -V2 = 5.629 x 1011

V2 = ( 1.7 x 106 )^2 - 5.629 x 1011

V = 1.525 x 106

So the velocity is 1.525 x 106 m/s

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