Question

Time taken to produce 1 mol of photon by a \(100 \mathrm{~W}\left(\mathrm{Js}^{-1}\right)\) yellow lamp is (Given \(\lambda\) of yellow light is \(500 \mathrm{~nm}\) ) : (1) \(340 \mathrm{~min}\) (2) 12 min (3) \(36 \mathrm{~min}\) (4) 40 min

Short answer

The time taken is approximately 40 minutes.

Step-by-step solution

Find the Energy of One Photon

Use the formula for the energy of a photon, which is given by Planck's equation: \[ E = \frac{hc}{\lambda} \]Where, - \( h \) is Planck's constant (\(6.626 \times 10^{-34} \) Js)- \( c \) is the speed of light (\(3 \times 10^8 \) m/s)- \( \lambda \) is the wavelength of the yellow light (\(500 \) nm or \(500 \times 10^{-9} \) m)Substitute the values:\[ E = \frac{(6.626 \times 10^{-34} \text{ Js}) \times (3 \times 10^8 \text{ m/s})}{500 \times 10^{-9} \text{ m}} = 3.98 \times 10^{-19} \text{ J} \]

Calculate the Energy Required to Produce 1 Mole of Photons

1 mole of photons contains Avogadro's number (\(N_A\)) of photons. Thus, energy required for 1 mole of photons is:\[ E_{\text{total}} = E \times N_A \]Where, - \( N_A \) is Avogadro’s number, \(6.022 \times 10^{23}\) mol⁻¹.Substitute the values:\[ E_{\text{total}} = (3.98 \times 10^{-19} \text{ J}) \times (6.022 \times 10^{23}) \approx 2.396 \times 10^{5} \text{ J/mol} \]

Calculate the Time Taken

Use the relationship between power, energy, and time: \[ P = \frac{E}{t} \]Rearrange to solve for time (\( t \)):\[ t = \frac{E}{P} \]Where- \( P \) is the power of the lamp (100 W or 100 J/s)- \( E \) is the energy required (\(2.396 \times 10^{5} \) J)Substitute the values:\[ t = \frac{2.396 \times 10^{5} \text{ J}}{100 \text{ J/s}} = 2396 \text{ s} \]

Convert Time into Minutes

Convert the time from seconds to minutes:\[ t = \frac{2396 \text{ s}}{60 \text{ s/min}} \approx 39.93 \text{ min} \]Rounding to two significant figures, we get approximately 40 min.

Key concepts

Photon Energy Calculation

Photons are tiny packets of light energy. To determine the energy of one photon, we use Planck's Equation:

• \[ E = \frac{hc}{\lambda} \]

Here,
- \( h \) is Planck's constant, valued at \(6.626 \times 10^{-34} \) Js.
- \( c \) is the speed of light, \(3 \times 10^8 \) meters per second.
- \( \lambda \), the wavelength of the yellow light, is \(500 \) nm (or \(500 \times 10^{-9} \) meters).
By plugging these values into the equation, we calculate the energy of one photon as:
\[ E = \frac{(6.626 \times 10^{-34} \text{ Js}) \times (3 \times 10^8 \text{ m/s})}{500 \times 10^{-9} \text{ m}} = 3.98 \times 10^{-19} \text{ J} \]
This energy value is extremely small, illustrating how tiny individual photon energies are. Each photon of yellow light carries about \( 3.98 \times 10^{-19} \) Joules of energy.

Planck's Constant

Planck's constant, denoted as \( h \), is a fundamental quantity in quantum mechanics. It acts as a proportional constant between the energy of photons and the frequency of their associated electromagnetic wave. The value of Planck's constant is \( 6.626 \times 10^{-34} \) Js.

• It is crucial in our energy calculation because it relates energy (E) to the frequency (f) of a photon via the equation: \[ E = hf \]
• This can also be expressed in terms of wavelength: \[ E = \frac{hc}{\lambda} \]


• Here, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the light.

Understanding Planck's constant helps bridge the gap between microscopic quantum phenomena and classical physics.
It enables the quantification of photon's energy, which is crucial for various applications like spectroscopy, calculating atomic and molecular binding energies, and understanding the behavior of electrons in atoms.

Avogadro's Number

Avogadro's number, denoted \( N_A \), is a constant that represents the number of units (atoms, molecules, or particles) in one mole of any substance. Its value is approximately \( 6.022 \times 10^{23} \) entities per mole. This concept is fundamental in chemistry and physics for converting between atomic-scale measurements and macroscopic quantities.

• When calculating the total energy needed to produce a mole of photons, we multiply the energy of one photon by Avogadro's number: \[ E_{\text{total}} = E \times N_A \]
• This helps us find the energy associated with macroscopic amounts of material from the microscopic energy of individual atoms or molecules.

For example, in the calculation provided, the energy of one photon (\( 3.98 \times 10^{-19} \text{ J} \)) is multiplied by Avogadro's number to find the energy required to produce one mole of photons (\( 2.396 \times 10^{5} \text{ J/mol} \)).
Such conversions are critical in fields like thermodynamics, kinetic theory of gases, and quantum chemistry, ensuring we can accurately describe and manipulate material quantities in scientific and industrial settings.