Hint: Insert equation 1 into equation 2, and simplify the resulting equation. In
ID: 856611 • Letter: H
Question
Hint: Insert equation 1 into equation 2, and simplify the resulting equation. Insert all the appropriate quantities into the formula to calculate a numerical value of t . Be sure to covert the mass of the electron into kg! The problems given below are due next week, but your group should work on them today. 2.(32 pts) Microwave radiation has a wavelength of 11.2 cm. (e) What is the frequency, in Hz, of this photon? (f) What is the wavenumber, in 1/cm, associated with the photon in (a) (g) What is the energy, in J, of a single photon of this radiation? (h) How many photons of this radiation are required to heat 200.0 mL of coffee from 22.0 degree C?Explanation / Answer
(e)
Given, wavelength = 11.2 cm.
frequency(nue) = c/lambda
C = speed of light = 3 x 10^8 m / s
lambda = 11.2 cm = 0.112 m
frequency = (3 x 10^8) / (0.112) = 2.68 x 10^9 s^-1
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(f)
Wavenumber = 1/wavelength = 1/lamba
Wavenumber = 1 / 11.2 = 0.089 cm^-1
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(g)
E = nhv,
where, n = no.of photons = 1
h = Planck's constant = 6.626 x 10^-34 J.s
v = frequency = 2.68 x 10^9 s^-1
E = (1) ( 6.626 x 10^-34 J.s) (2.68 x 10^9 s^-1)
E = 1.77 x 10^-24 J
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(h)
First, we need to find the heat when 200.0 mL coffee is heated from 22 degree C to 65 degree C.
Q = mcdelta T
let us consider the density of coffee and specific heat are equal to the density, and specific heat of water.
m = 200.0 g
c = 4.184 j/g oC
delta T = Tfinal - Tinitial = 65 - 22 = 43 degree C
Q = 200 x 4.184 x 43
Q = 35982.4 J
Q = 35.98 kJ
So, this much heat is required.
So, in equation E = nhv,
E = 35982.4 J
n = ?
hv = 1.77 x 10^-24 J
n = E / hv
n = 2.033 x 10^28 photons are required.
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