Problem 8 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

A bulb emits light of wavelength \(\frac{1987.8}{7} \mathrm{~nm} .\) The bulb is rated as \(200 \mathrm{~W}\) and \(14 \%\) of the energy is emitted as light. How many photons are emitted by the bulb per second? (a) \(1.2 \times 10^{21}\) (b) \(4 \times 10^{10}\) (c) \(1.33 \times 10^{11}\) (d) \(4 \times 10^{19}\)

Step-by-Step Solution

Answer

The number of photons emitted by the bulb per second is approximately \(1.33 \times 10^{19}\), which aligns with option (d).

Step 1: Calculate the energy emitted as light per second

First determine the total energy emitted as light per second by the bulb, knowing that 14% of the bulb's power is converted to light. The bulb's power is 200W, which means the energy emitted as light per second is: \( 0.14 \times 200W = 28W \).

Step 2: Determine the energy of a single photon

Calculate the energy of a single photon using the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, Js) \), \( c \) is the speed of light \( (3 \times 10^{8} \, m/s) \), and \( \lambda \) is the wavelength of light \( (\frac{1987.8}{7} \, nm) \). Convert the wavelength from nanometers to meters by multiplying by \( 10^{-9} \).

Step 3: Calculate the number of photons emitted per second

To find the number of photons emitted per second, divide the total energy emitted per second by the energy of a single photon: \( \text{number of photons} = \frac{\text{total energy per second}}{\text{energy of a single photon}} \).

Step 4: Perform the calculations

First, calculate the energy of a single photon, then calculate the number of photons by the division method outlined above. Convert all units properly to ensure coherency in the calculations.

Key Concepts

Planck's Constant

Understanding Planck's constant is crucial when studying quantum physics and photon emission calculations. It's a fundamental constant denoted by the symbol 'h', with a value of approximately \( 6.626 \times 10^{-34} \text{ Js} \). This seemingly small number is rather powerful, as it relates to the quantization of energy in the realm of the very small, such as electrons and photons.

The constant was discovered by Max Planck, one of the pioneers of quantum mechanics. His groundbreaking work revealed that energy is not continuous, but rather comes in discrete 'packets' called quanta. Planck's constant is a measure of these packets. In the world of photon emission, when an atom's electron drops to a lower energy level, it releases energy in the form of a photon, and the amount of energy is directly tied to Planck’s constant.

For instance, to compute the energy of a photon we use the formula \( E = h u \) where \( u \) is the frequency of the photon. It's this relationship that allows us to tackle problems involving photon emissions, such as calculating the number of photons emitted by a light source.

Energy of a Photon

The energy of a photon is defined by its frequency and Planck's constant. The formula \( E = \frac{hc}{\lambda} \) is used to calculate this energy, with 'E' representing the energy of a photon, 'h' standing for Planck's constant, 'c' for the speed of light in a vacuum (\(3 \times 10^{8} \text{ m/s}\)), and '\(\lambda\)' for the photon's wavelength.

Photons are massless particles and are the basic units of light with dual properties, behaving as both waves and particles. Their energy is a crucial factor, as it determines the color of light in the visible spectrum and the type of electromagnetic radiation, ranging from gamma rays to radio waves with differing application and effects.

The amount of energy carried by a photon decides how much it can interact with other particles or objects. For example, in medical applications, high-energy photons in the form of X-rays can penetrate the body, but lower-energy photons in visible light cannot.

Wavelength and Frequency Relationship

The wavelength and frequency of a photon are inversely related through the speed of light. This relation is encapsulated in the formula \( c = \lambdau \) where 'c' is the speed of light, '\(\lambda\)' is the wavelength, and '\(u\)' is the frequency. If the wavelength is given, this mathematical relationship can be used to find the frequency and vice versa.

In the context of the bulb problem, the given wavelength can be utilized to determine the frequency of the emitted light. Since the speed of light is a constant, a higher frequency implies a shorter wavelength and more energy per photon. The frequency is particularly important because it directly affects the energy of a photon, which is essential to calculate the number of photons emitted.

It's also worth noting that this relationship is fundamental to understanding the electromagnetic spectrum. For example, microwaves have a longer wavelength and thus lower frequency compared to visible light, resulting in their varied characteristics and applications.