Answer to Question #97826 in General Chemistry for Brittany Wallace

Question

1. For hydrogen, which of the following electron transitions emits the longest wavelength of light?
A. n = 5 to n = 4
B. n = 4 to n = 1
C. n = 4 to n = 2
D. n = 3 to n = 2
E. n = 2 to n = 1

2. Calculate the energy of a photon released by the n = 7 to n = 3 transition in the hydrogen emission spectrum.

3. Calculate the de Broglie wavelength, in meters, of a 5500-lb sport-utility vehicle moving at 57 miles per hour.

Answer 1

1.from Rydberg Formula we know 1/wavelength proportional to (1/n₁² - 1/n₂² )

thus from the above

wave length will be highest for from n = 5 to n = 4

2.energy of emission from n = 7 to n = 3

according to Rydberg formula 1/ wavelength = z * R (1/n₁² - 1/n₂² )

where z = atomic number = 1

R = Rydberg constant = 1.09737×10⁷ m⁻¹

1/wavelength = 995346.93 m^-1

energy = hc * (1/wavelength) h = planck constant = 6.626 *10^-34 m² kg s^-1

c = 3* 10⁸ m s^-1

energy will be equal to 1.9 * 10^-19 joule

3.weight of the veichle = 5500 lb = 2494.75 kg

speed 57 miles per hour = 25.48 m/s

thus momentum = 2494.75 * 25.48 = 63566.23 kg m s^-1

thus De broglie wavelength = 1.04237 * 10 ^{- 38} m