Question

question 35 (mandatory) (1 point)
listen
which of the following photons has the highest energy?
a) a photon with v = 1.55 × 10¹⁵ hz
b) a photon with λ = 633 pm
c) a photon with λ = 1.064 μm
d) a photon with v = 5.83 × 10¹⁴ hz
e) a photon with λ = 647 nm

Explanation:

Step1: Recall the energy formula for photons

The energy of a photon is given by \(E = h
u\), where \(h = 6.626\times10^{-34}\space J\cdot s\) is Planck's constant and \(
u\) is the frequency. Also, from \(c=\lambda
u\) (\(c = 3\times10^{8}\space m/s\)), we can get \(
u=\frac{c}{\lambda}\).

Step2: Calculate the frequency for option b

Given \(\lambda = 633\space pm=633\times 10^{-12}\space m\). Then \(
u=\frac{c}{\lambda}=\frac{3\times 10^{8}}{633\times 10^{-12}}\space Hz\approx4.74\times 10^{17}\space Hz\)

Step3: Calculate the frequency for option c

Given \(\lambda = 1.064\space\mu m = 1.064\times 10^{-6}\space m\). Then \(
u=\frac{c}{\lambda}=\frac{3\times 10^{8}}{1.064\times 10^{-6}}\space Hz\approx2.82\times 10^{14}\space Hz\)

Step4: Calculate the frequency for option e

Given \(\lambda = 647\space nm=647\times 10^{-9}\space m\). Then \(
u=\frac{c}{\lambda}=\frac{3\times 10^{8}}{647\times 10^{-9}}\space Hz\approx4.64\times 10^{14}\space Hz\)

Step5: Compare the frequencies

Option a: \(
u = 1.55\times 10^{15}\space Hz\)
Option b: \(
u\approx4.74\times 10^{17}\space Hz\)
Option c: \(
u\approx2.82\times 10^{14}\space Hz\)
Option d: \(
u = 5.83\times 10^{14}\space Hz\)
Option e: \(
u\approx4.64\times 10^{14}\space Hz\)

Since \(E = h
u\) and \(h>0\), the photon with the highest frequency has the highest energy. Among the calculated frequencies, the frequency of option b (\(
u\approx4.74\times 10^{17}\space Hz\)) is the highest.

Answer:

B. a photon with \(\lambda = 633\space pm\)