1) A) An electron in the n = 4 level of the hydrogen atom relaxes to a lower ene

Question

1) A) An electron in the n = 4 level of the hydrogen atom relaxes to a lower energy level, emitting light of = 97.3 nm . Find the principal level to which the electron relaxed.

   B) Determine the wavelength of the light absorbed when an electron in a hydrogen atom makes a transition from an orbital in which n = 2 to an orbital in which n =5.

C) A 0.22-caliber handgun fires a 2.9-g bullet at a velocity of 775 m/s. Calculate the de Broglie wavelength of the bullet.

D) An electron has a de Broglie wavelength of 225 nm. What is the speed of the electron?

Thank you so much in advance!

Explanation / Answer

A) 1/ = R (1/n12 - 1/n22)

or, 1/(97.3 *10-7 cm) = (109737 cm-1) ( 1/ n12 - 1/42)

or, n1 = 1

B) 1/ = R (1/n12 - 1/n22)

= (109737 cm-1) ( 1/ 22 - 1/52)

= 23044.77 cm-1

or, = 4339 angstrom

C. = h/mv

= (6.626*10-31 J.s) / (0.0029 kg)(775 m/s)

= 2.94*10-31 m

D. speed of the electron, v = h/m

= (6.626*10-31 J.s) / (9.1*1031 kg)( 225 * 10-9 m)

= 3.236 * 106 m/s

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