Q.The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz (kilo hertz). Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to?

The key idea is that every spectral line in hydrogen corresponds to a transition between two energy levels. The Lyman series involves transitions that end at $n=1$. The energy of the photon emitted equals the difference between the initial and final level energies. Once we know which level the electron started from, we can compute the energy needed to jump to the next higher level.

The energy of an electron in the $n$th orbit of hydrogen is

The Lyman line in question comes from a transition $n_x \to 1$, so its energy is

We are told this energy is $1.635 \times 10^{-18}$ J. Let's find $n_x$.

1. Set up the equation

2. Divide through by $2.18 \times 10^{-18}$

3. Solve for $1/n_x^2$

Hence $n_x^2 = 4$, so $n_x = 2$.

So the electron is in the $n=2$ state. The question asks for the energy required to excite it from $n=2$ to $n=3$.

4. Energy of $n=2$ state

5. Energy of $n=3$ state

6. Excitation energy

This rounds to $3 \times 10^{-19}$ J.

The photoelectric effect describes the emission of electrons from a metal surface when light shines on it. This phenomenon is governed by the energy relationship between the incident light and the metal's properties.

The key concept here is that light consists of discrete packets of energy called photons. For an electron to be ejected from a metal, it must absorb a photon with sufficient energy to overcome the forces binding it to the metal. This minimum energy required is known as the work function ($W_0$) of the metal.

For photoemission to occur, the energy of the incident photon ($E$) must be greater than or equal to the work function ($W_0$) of the metal. If $E < W_0$, no photoelectrons will be emitted, regardless of the intensity of the light.

We need to calculate the energy of the photons corresponding to the given wavelength and then compare this energy with the work functions of metals A, B, and C.

1. Calculate the energy of the incident photons:

   The wavelength of the incident light is given as $\lambda = 400$ nm. We first convert this to meters:

   $\lambda = 400 \times 10^{-9}$ m

   Now, we use the formula for photon energy:

   $E = \frac{hc}{\lambda}$

   Substitute the given values for Planck's constant ($h$), the speed of light ($c$), and the wavelength ($\lambda$):

   $E = \frac{(6.6 \times 10^{-34} \text{ Js}) \times (3 \times 10^8 \text{ ms}^{-1})}{400 \times 10^{-9} \text{ m}}$

   $E = \frac{19.8 \times 10^{-26}}{400 \times 10^{-9}}$ J

   $E = \frac{19.8}{400} \times 10^{-17}$ J

   $E = 0.0495 \times 10^{-17}$ J

   $E = 4.95 \times 10^{-19}$ J

   To compare this energy with the work functions, which are given in electron volts (eV), we convert the photon energy from Joules to eV using the conversion factor $1$ eV $= 1.6 \times 10^{-19}$ J:

   $E_{\text{eV}} = \frac{4.95 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}}$

   $E_{\text{eV}} = \frac{4.95}{1.6}$ eV

   $E_{\text{eV}} = 3.09375$ eV

   For practical comparison, we can use $E \approx 3.09$ eV.

2. Compare the photon energy with the work functions of metals A, B, and C:

   We now check the condition for photoelectron emission ($E > W_0$) for each metal:

   • Metal A: Work function $W_{0A} = 2.25$ eV

     Since $E = 3.09$ eV and $W_{0A} = 2.25$ eV, we have $E > W_{0A}$ ($3.09 \text{ eV} > 2.25 \text{ eV}$).

     Therefore, photoelectrons will be emitted from metal A.

   • Metal B: Work function $W_{0B} = 2.42$ eV

     Since $E = 3.09$ eV and $W_{0B} = 2.42$ eV, we have $E > W_{0B}$ ($3.09 \text{ eV} > 2.42 \text{ eV}$).

     Therefore, photoelectrons will be emitted from metal B.

   • Metal C: Work function $W_{0C} = 3.7$ eV

     Since $E = 3.09$ eV and $W_{0C} = 3.7$ eV, we have $E < W_{0C}$ ($3.09 \text{ eV} < 3.7 \text{ eV}$).

     Therefore, photoelectrons will NOT be emitted from metal C.

3. Identify the metals from which photoelectrons are emitted:

   Based on the comparisons, photoelectrons are emitted from metals A and B.

The concept here is the Bohr model of the hydrogen atom, which gives a simple formula for the radius of an electron orbit:

where $a_0 = 52.9\ \text{pm}$ (the Bohr radius) and $n$ is the principal quantum number. Each spectral series corresponds to transitions ending at a specific $n_f$:

• Lyman series: $n_f = 1$
• Balmer series: $n_f = 2$
• Paschen series: $n_f = 3$
• Pfund series: $n_f = 5$

So if we can find $n_f$ from the given radius, we immediately know the series.

1. Write the radius formula The radius of the $n$-th orbit in hydrogen is:

2. Plug in the given radius The problem states the final orbit has radius $476.1\ \text{pm}$. So:

3. Solve for $n_f^2$

Compute: $476.1 \div 52.9 \approx 9.00$ (since $52.9 \times 9 = 476.1$ exactly).

4. Find $n_f$

5. Identify the series Transitions ending at $n_f = 3$ belong to the Paschen series.

The question asks us to identify the correct statements regarding properties of light and atomic spectra. We need to evaluate each statement based on fundamental principles of optics and atomic physics.

Concept and Intuition

1. Electromagnetic Spectrum: Light is an electromagnetic wave, characterized by its wavelength ($\lambda$) and frequency ($f$). These are related by the speed of light ($c$) in a vacuum: $c = f\lambda$. In the visible spectrum, different colors correspond to different wavelengths and frequencies.
2. Dispersion of Light: When white light passes through a medium like a prism, it splits into its constituent colors. This phenomenon, called dispersion, occurs because the refractive index of the medium varies with the wavelength of light. Shorter wavelengths (like violet) generally experience a higher refractive index and thus greater deviation than longer wavelengths (like red).
3. Atomic Spectra: Atoms, particularly in the gas phase, emit light when their electrons transition between discrete energy levels. Because these energy levels are quantized, the emitted light consists of specific, discrete wavelengths, forming a line spectrum rather than a continuous spectrum.

Let's evaluate each statement:

1. Statement I: In the visible spectrum, violet light has highest frequency and red light has lowest frequency.

   • The visible spectrum ranges approximately from 400 nm (violet) to 700 nm (red).
   • The relationship between the speed of light ($c$), frequency ($f$), and wavelength ($\lambda$) is given by:

     $c = f\lambda$

   • From this, we can express frequency as $f = c/\lambda$.
   • Since $c$ is a constant, frequency is inversely proportional to wavelength.
   • Violet light has the shortest wavelength in the visible spectrum (around 400 nm), so it will have the highest frequency.
   • Red light has the longest wavelength in the visible spectrum (around 700 nm), so it will have the lowest frequency.
   • Therefore, Statement I is correct.

2. Statement II: When white light is passed through a prism, violet light is deviated the most and red light is deviated the least.

   • When white light passes through a prism, it undergoes dispersion. This means different colors (wavelengths) of light are refracted by different amounts.
   • The refractive index ($n$) of a material generally decreases as the wavelength ($\lambda$) of light increases. This phenomenon is known as normal dispersion.
   • Violet light has a shorter wavelength than red light. Consequently, the refractive index of the prism material for violet light ($n_v$) is greater than for red light ($n_r$).
   • The deviation ($\delta$) produced by a prism is directly related to its refractive index. For a given prism angle, a higher refractive index leads to greater deviation.
   • Since $n_v > n_r$, violet light will be deviated more than red light.
   • Therefore, Statement II is correct.
   Tip

   Remember the acronym VIBGYOR for the visible spectrum. As you move from Violet to Red, wavelength increases, frequency decreases, and deviation by a prism decreases.

3. Statement III: Hydrogen atoms in gas phase exhibit line spectrum.

   • According to the Bohr model and quantum mechanics, electrons in atoms can only exist in specific, discrete energy levels.
   • When an electron in an excited atom transitions from a higher energy level to a lower one, it emits a photon. The energy of this photon is exactly equal to the difference between the two energy levels.
   • Since the energy levels are discrete, the energy differences are also discrete. This means the emitted photons will have specific, discrete energies, and thus specific, discrete frequencies and wavelengths ($E = hf = hc/\lambda$).
   • When these specific wavelengths are observed, they appear as distinct bright lines against a dark background, forming a line emission spectrum.
   • Hydrogen atoms, when excited in the gas phase (e.g., in a discharge tube), are a classic example of this phenomenon, exhibiting characteristic line spectra (like the Balmer series in the visible region).
   • Therefore, Statement III is correct.

Since all three statements (I, II, and III) are correct, the option that includes all of them is the correct answer.

The key concept is Bragg’s law, which relates the wavelength of X-rays to the spacing between crystal planes and the angle at which constructive interference (a diffraction peak) occurs. The law is:

where:

• $n$ is the order of reflection (given as 1),
• $\lambda$ is the wavelength,
• $d$ is the interplanar spacing,
• $\theta$ is the glancing angle — the angle between the incident X-ray beam and the crystal plane.

A classic pitfall: the problem gives $2\theta = 60^\circ$, but Bragg’s law uses $\theta$, not $2\theta$. So $\theta = 30^\circ$.

Now, step by step:

1. Identify the glancing angle The diffraction peak occurs at $2\theta = 60^\circ$. In X-ray diffraction, $2\theta$ is the angle between the incident and diffracted beams. The angle used in Bragg’s law is $\theta$, the angle the beam makes with the crystal plane. Hence:

2. Write Bragg’s law with given values We have $n = 1$, $d = 1.54\ \text{Å}$, and $\sin 30^\circ = 0.5$. So:

3. Convert the wavelength to centimetres The answer choices are in cm. Recall:

Therefore:

4. Match with the options Option (D) is $1.54 \times 10^{-8}$, which matches exactly. Option (A) is $1.54$ (no exponent, so in Å, not cm). Options (B) and (C) have wrong exponents.

The key to this question lies in understanding nuclear stability. Elements are radioactive when their nuclei are unstable and spontaneously emit radiation to reach a more stable configuration. For most lighter elements, stable isotopes exist alongside any radioactive ones. But as atomic number increases, the proton-proton repulsion in the nucleus grows so strong that no arrangement of neutrons can produce a stable configuration.

Radon sits at atomic number 86, well into the region where nuclear forces can no longer hold nuclei together indefinitely. Every single isotope of radon is radioactive—there are no stable forms whatsoever.

Let me walk through each noble gas:

1. Argon (Ar, Z = 18) has three naturally occurring isotopes: $^{36}\text{Ar}$, $^{38}\text{Ar}$, and $^{40}\text{Ar}$. The first two are completely stable. $^{40}\text{Ar}$ is the dominant isotope (99.6% abundance) and is also stable, though it happens to be the decay product of radioactive $^{40}\text{K}$.

2. Krypton (Kr, Z = 36) has six stable isotopes ($^{78}\text{Kr}$, $^{80}\text{Kr}$, $^{82}\text{Kr}$, $^{83}\text{Kr}$, $^{84}\text{Kr}$, $^{86}\text{Kr}$) that make up natural krypton. While radioactive isotopes of krypton exist (like $^{85}\text{Kr}$ from nuclear fission), the element in its natural form is predominantly stable.

3. Xenon (Xe, Z = 54) has nine stable isotopes. Natural xenon is a mixture of these stable forms, though trace amounts of radioactive $^{127}\text{Xe}$ can appear from cosmic-ray interactions.

4. Radon (Rn, Z = 86) has no stable isotopes at all. The most common natural isotope is $^{222}\text{Rn}$ (half-life 3.8 days), produced in the uranium-238 decay chain. Other isotopes like $^{220}\text{Rn}$ (from thorium-232 decay) and $^{219}\text{Rn}$ (from uranium-235 decay) also occur naturally, but all are radioactive.

The pattern is clear: as we move down the noble gas group, nuclear instability increases. Radon crosses the threshold where stability becomes impossible.