Problem 10 - Master Resource Book in JEE Main Physics

Question

An important spectral emission line has a wavelength of \(21 \mathrm{~cm}\). The corresponding photon energy is \(\left(h=6.62 \times 10^{-34}\right.\) \(\begin{array}{ll}\text { (a) } 5.9 \times 10^{-8} \mathrm{eV} & \text { (b) } 5.9 \times 10^{-4} \mathrm{eV}\end{array}\) (c) \(5.9 \times 10^{-6} \mathrm{eV}\) (d) \(11.8 \times 10^{-6} \mathrm{eV}\)

Step-by-Step Solution

Answer

The corresponding photon energy is option (c), \(5.9 \times 10^{-6} \text{ eV}\).

Step 1: Convert wavelength to meters

The given wavelength is 21 cm. First, convert this wavelength into meters to use it in further calculations.Wavelength in meters: \[21 \text{ cm} = 21 \times 10^{-2} \text{ m} = 0.21 \text{ m}\]

Step 2: Use the energy formula for photons

To find the energy of the photon, use the formula:\[E = \frac{hc}{\lambda}\]where:- \(h = 6.62 \times 10^{-34} \text{ J s}\) is the Planck's constant,- \(c = 3.00 \times 10^{8} \text{ m/s}\) is the speed of light,- \(\lambda = 0.21 \text{ m}\) is the wavelength.

Step 3: Calculate photon energy

Substitute the known values into the formula:\[E = \frac{6.62 \times 10^{-34} \times 3.00 \times 10^{8}}{0.21}\]Calculate:\[E = \frac{1.986 \times 10^{-25}}{0.21} = 9.457 \times 10^{-25} \text{ J}\]

Step 4: Convert energy to electronvolts

To convert the energy from joules to electronvolts, use the conversion factor \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\):\[E = \frac{9.457 \times 10^{-25} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}}\]Calculate:\[E = 5.9 \times 10^{-6} \text{ eV}\]

Step 5: Find the correct option

Looking at the calculated value of the energy, it matches option (c) which is \(5.9 \times 10^{-6} \text{ eV}\). This is the energy corresponding to the photon's wavelength.

Key Concepts

Planck's constant

Planck's constant is an essential element in quantum mechanics, a field of physics that studies the behavior of extremely small particles like photons and electrons. It represents the proportionality factor between the energy

• Energy, represented as \( E \)
• Frequency, represented as \( u \)

It is often denoted by \( h \) and has a value of approximately \( 6.62 \times 10^{-34} \) Joule seconds (J s). This small constant is crucial because it helps determine the energy levels that photons can possess.

In the formula \( E = h u \) , Planck’s constant \( h \) connects energy with frequency, allowing for the calculation of photon energy if the frequency is known.

In problems related to light and electromagnetic radiation, such as converting wavelength to energy, understanding Planck’s constant is essential for accurate calculations.

Wavelength to meters conversion

Converting a wavelength from centimeters to meters is a simple yet crucial part of any calculation involving physical constants, like those that arise in studying electromagnetic radiation.

• Wavelength is commonly given in different units like nanometers, centimeters, or meters.
• In science, using the SI unit, meters, is often preferred for consistency.

For example, to convert a given wavelength of 21 centimeters into meters:
21 cm can be expressed as \( 21 \times 10^{-2} \) m, which simplifies to 0.21 m.

This conversion is needed when using formulas such as \( E = \frac{hc}{\lambda} \), where \( \lambda \) denotes the wavelength in meters. Getting this right ensures the calculations remain accurate leading to a correct determination of energy.

Energy conversion to electronvolts

Converting energy from joules to electronvolts (eV) is an important step when working with photons, especially since electronvolts are a more convenient unit of energy at atomic scales.

• 1 eV is equivalent to \( 1.602 \times 10^{-19} \) Joules.

This conversion is useful because it allows physicists and students alike to more intuitively grasp the energy values, since energies in electronvolts often fall into more manageable numerical ranges.

For instance, if a photon’s energy is calculated as \( 9.457 \times 10^{-25} \) Joules, converting this to electronvolts involves dividing by the conversion factor:
\[ E = \frac{9.457 \times 10^{-25} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} = 5.9 \times 10^{-6} \text{ eV} \].
Understanding this conversion technique ensures you can seamlessly switch between units and interpret findings correctly, especially in exams or practical applications where the electronvolt is more applicable.