Question

a. Starting with the expression$\Delta f\Delta t\approx 1$for a wave packet, find an expression for the product

$\Delta E\Delta t$for a photon.

b. Interpret your expression. What does it tell you?

c. The Bohr model of atomic quantization says that an atom in an excited state can jump to a lower-energy state by emitting a photon. The Bohr model says nothing about how long this process takes. You'll learn in Chapter 41 that the time any particular atom spends in the excited state before emitting a photon is unpredictable, but the average lifetime $\Delta t$of many atoms can be determined. You can think of $\Delta t$as being the uncertainty in your knowledge of how long the atom spends in the excited state. A typical value is $\Delta t\approx 10\mathrm{ns}$. Consider an atom that emits a photon with a $500\mathrm{nm}$wavelength as it jumps down from an excited state. What is the uncertainty in the energy of the photon? Give your answer in $\mathrm{eV}$.

d. What is the fractional uncertainty $\Delta E/E$in the photon's energy?

Short answer

a.) $\Delta E\Delta t\approx h$

b.) We can't know the exact energy of a photon, where the uncertainty in the knowledge of a photon's energy depends on the time period taken to measure this energy.

c.) $\Delta E=4.14\times {10}^{-7}\mathrm{eV}$

d.)$\frac{\Delta E}{E}=1.67\times {10}^{-7}$

Step-by-step solution

a.) Given Information : Expression for a wave packet ΔfΔt≈1

a.) We know that the energy of a photon is E = hf, which means that $\Delta E=h\Delta f$. Now, we have been told in the question that the expression $\Delta f\Delta t\approx 1$is valid, meaning that we can treat the photon as a wave packet. Substitute $\Delta E/h$for $\Delta f$

$\frac{\Delta E}{h}\times \Delta t\approx 1$

$\Delta E\Delta t\approx h$

b.) Given Information : Interpretation on the basis of part a 

b.) We can't know the exact energy of a photon, where the uncertainty in the knowledge of a photon's energy depends on the time period taken to measure this energy.

c.) Given Information : an atom that emits a photon with a  500 nm wavelength as it jumps down from an excited state. 

We can simply use the equation obtained in part (a) to find $\Delta E$when $\Delta t=10\mathrm{ns}$

$\Delta E=\frac{h}{\Delta t}=\frac{6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}}{10\times {10}^{-9}\mathrm{s}}=6.626\times {10}^{-26}\mathrm{J}$

$\Delta E=6.626\times {10}^{-26}\mathrm{J}\times \frac{1\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}}=4.14\times {10}^{-7}\mathrm{eV}$

d.) Given Information : The photon's energy is

$E=\frac{hc}{\lambda }=\frac{\left(6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)}{500\times {10}^{-9}\mathrm{m}}=3.976\times {10}^{-19}\mathrm{J}$

$E=3.976\times {10}^{-19}\mathrm{J}\times \frac{1\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}}=2.485\mathrm{eV}$

the fractional uncertainty is

$\frac{\Delta E}{E}=\frac{4.14\times {10}^{-7}\mathrm{eV}}{2.485\mathrm{eV}}=1.67\times {10}^{-7}$