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practice problems 1. a photon has a frequency (ν) of 2.68 × 10⁶ hz. cal…

Question

practice problems

  1. a photon has a frequency (ν) of 2.68 × 10⁶ hz. calculate its energy.

ans: e = 1.78 × 10⁻²⁷ j

  1. calculate the energy (e) and wavelength (λ) of a photon of light with a frequency (ν) of 6.165 × 10¹⁴ hz.

ans: e = 4.1 × 10⁻¹⁹ j λ = 4.87 × 10⁻⁷ m

  1. calculate the frequency and the energy of blue light that has a wavelength of 400 nm (h = 6.62 × 10⁻³⁴ j·s).

ans: ν = 7.5 × 10¹⁴ hz e = 4.97 × 10⁻¹⁹ j

  1. calculate the wavelength and energy of light that has a frequency of 1.5 × 10¹⁵ hz.

ans: λ = 2.0 × 10⁻⁷ m e = 9.95 × 10⁻¹⁹ j

  1. a photon of light has a wavelength of 0.050 cm. calculate its energy.

ans: e = 3.98 × 10⁻²² j

  1. calculate the number of photons having a wavelength of 10.0 μm required to produce 1.0 kj of energy.

ans: 5.0 × 10²² photons

  1. calculate the total energy in 1.5 × 10¹³ photons of gamma radiation having λ = 3.0 × 10⁻¹² m.

ans: 1.0 j

  1. calculate the energy and frequency of red light having a wavelength of 6.80 × 10⁻⁵ cm.

ans: e = 2.92 × 10⁻¹⁹ j ν = 4.4 × 10¹⁴ hz

  1. the wavelength of green light from a traffic signal is centered at 5.20 × 10⁻⁵ cm. calculate the frequency.

ans: λ = 5.77 × 10¹⁴ hz.

  1. calculate the frequency of light that has a wavelength of 4.25 × 10⁻⁹ m. identify the type of electromagnetic radiation.

ans: ν = 7.1 × 10¹⁶ hz. uv radiation

equations and constants:
e = hν and e = hc/λ
c = λν c/λ = ν
e = energy of one photon with a frequency of ν
c = speed of light = 3.0 × 10⁸ m/s (meters per second)
h = planck’s constant = 6.63 × 10⁻³⁴ j·s
λ = wavelength in meters
ν = frequency in hz (waves/s or 1/s or s⁻¹)

Explanation:

Response

Let's solve problem 1 as an example (since the user might want to solve one of these photon energy/frequency/wavelength problems, we'll pick problem 1: "A photon has a frequency (ν) of 2.68 × 10⁶ Hz. Calculate its energy.").

Step 1: Recall the formula for photon energy

The energy \( E \) of a photon is given by \( E = h
u \), where \( h \) is Planck's constant (\( h = 6.63 \times 10^{-34} \, \text{J-s} \)) and \(
u \) is the frequency.

Step 2: Substitute the given values into the formula

We know \(
u = 2.68 \times 10^{6} \, \text{Hz} \) and \( h = 6.63 \times 10^{-34} \, \text{J-s} \). Plugging these into \( E = h
u \):

\( E = (6.63 \times 10^{-34} \, \text{J-s}) \times (2.68 \times 10^{6} \, \text{Hz}) \)

Step 3: Multiply the numbers and the exponents

First, multiply the coefficients: \( 6.63 \times 2.68 \approx 17.77 \)

Then, multiply the exponents using the rule \( a^m \times a^n = a^{m + n} \): \( 10^{-34} \times 10^{6} = 10^{-34 + 6} = 10^{-28} \)

So, \( E \approx 17.77 \times 10^{-28} \, \text{J} \)

Step 4: Rewrite in proper scientific notation

\( 17.77 \times 10^{-28} = 1.777 \times 10^{-27} \, \text{J} \) (which matches the given answer \( 1.78 \times 10^{-27} \, \text{J} \) when rounded appropriately)

Answer:

The energy of the photon is \( \boldsymbol{1.78 \times 10^{-27} \, \text{J}} \) (or approximately \( 1.777 \times 10^{-27} \, \text{J} \) before rounding).

If you want to solve a different problem (e.g., problem 2, 3, etc.), please specify, and we can follow a similar step-by-step process using the relevant formulas (\( E = h
u \), \( c = \lambda
u \), \( E = \frac{hc}{\lambda} \)) and the given constants (\( h = 6.63 \times 10^{-34} \, \text{J-s} \), \( c = 3.0 \times 10^{8} \, \text{m/s} \)).