A $100\ \text{W}$ sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is $589\ \text{nm}$. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?

A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large s

Exercises · 11.7

100\ \text{W}

(a) What is the energy per photon associated with the sodium light?

(b) At what rate are the photons delivered to the sphere?

Gujarat GsebTextbookSubjective· 3mImportance★★★★★

12% · 10/83 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Photon Energy

Photon Energy: The Energy Carried by Light

Light does not carry its energy in a continuous stream. It comes in discrete packets — like individual drops instead of a steady hose. Each packet is called a photon: the smallest possible unit of light of a given frequency. You cannot have half a photon; it is all or nothing.

The Intuition: Colour Determines Energy

Red light and blue light are both light, but they behave differently — blue light causes sunburn and drives chemical reactions more readily than red. The reason is that the colour (frequency) of light directly fixes how much energy each photon carries, and blue photons carry more energy than red photons.

The Planck–Einstein Relation

The energy $E$ of a single photon is directly proportional to its frequency $f$ , and inversely proportional to its wavelength $\lambda$ :

E = h f = \frac{h c}{\lambda}

Where:

$E$ = energy of one photon (joules, J)
$f$ = frequency (hertz, Hz)
$\lambda$ = wavelength (metres, m)
$c$ = speed of light in vacuum ( $3.00 \times 10^8$ m/s)
$h$ = Planck's constant ( $6.626 \times 10^{-34}$ J·s)

Important

Planck's constant $h$ is extremely tiny, so an individual photon carries a very small amount of energy — which is why we never feel single photons striking our skin.

What This Tells You

Higher frequency = higher energy. Gamma rays have extremely energetic photons; radio waves have very low-energy photons.
Shorter wavelength = higher energy. Ultraviolet photons are more energetic than visible-light photons; infrared photons are less.
Energy is quantised. Light energy comes in fixed packets — you can have 1, 2, or 1000 photons, but never 0.5. This was Max Planck's revolutionary idea of 1900.

This is why UV light causes sunburn (UV photons carry enough energy to damage DNA, while visible photons do not), why X-ray photons are energetic enough to pass through soft tissue but get absorbed by bone, and why a photon needs enough energy to actually break a bond or drive a photochemical/biological reaction, rather than merely wavelength-dependent intensity.

A Concrete Example

Note

Question: Which has more energy — a photon of red light ( $\lambda = 700$ nm) or a photon of blue light ( $\lambda = 450$ nm)?

Since $E = \dfrac{hc}{\lambda}$ , energy is inversely proportional to wavelength, so the shorter-wavelength blue photon carries more energy:

Red: $E_{\text{red}} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{700 \times 10^{-9}} \approx 2.84 \times 10^{-19}$ J …

Why this formula?

Why Photon Energy is $E = hf$ — The Reasoning Behind the Formula

The formula $E = hf$ is not something you memorise and plug numbers into. It comes from a deep shift in how physicists understood light.

The problem that forced the formula

By the late 1800s, classical physics said light was a wave — continuous, carrying energy in proportion to its intensity (brightness). But blackbody radiation and the photoelectric effect showed something bizarre: when you shine light on a metal, electrons are ejected only if the light's frequency is above a certain threshold, no matter how bright the light is. Below that frequency, even the brightest lamp won't kick out a single electron. Classically this made no sense — a wave's energy depends on amplitude, not frequency.

Einstein's radical idea (1905)

Einstein proposed that light comes in discrete packets — photons — each carrying a fixed energy that depends only on its frequency, with Planck's constant $h$ as the proportionality:

A single photon's energy is $E = hf$ , where $f$ is the frequency of the light.

This directly explains the photoelectric effect: an electron needs a minimum energy (the work function $\phi$ ) to escape. If a photon's energy $hf < \phi$ , no electron is ejected — regardless of how many photons you send, because each photon interacts with one electron individually.

Important

$E = hf$ is not derived from deeper principles — it is a postulate justified by the experiments it explains, and it unified optics with quantum mechanics.

The equivalent forms

Since $c = f\lambda$ for light in vacuum, the same idea in terms of wavelength is:

E = \frac{hc}{\lambda}

Use this when you're given $\lambda$ instead of $f$ . A photon also carries momentum (despite having no mass): …

Start your 14-day free trial to unlock the full solution →

Compute the energy of one 589 nm sodium photon from $E=hc/\lambda$ (≈2.11 eV), then divide the lamp's 100 W output by this to get the photon delivery rate (≈2.96×10^20 photons/s), since the sphere absorbs the lamp's entire output.

Step 1 — Energy per photon.

E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3\times10^{8})}{589\times10^{-9}} = \frac{1.989\times10^{-25}}{5.89\times10^{-7}}

E \approx 3.38\times10^{-19}\ \text{J}

Converting to electron-volts:

E = \frac{3.38\times10^{-19}}{1.6\times10^{-19}} \approx 2.11\ \text{eV}

Step 2 — Photon delivery rate. …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. Monochromatic light of frequency 6.0 x 10^14 Hz is produced by a laser. The power emitted is 2.0 x 10^-3 W. How many photons per second on an average, are emitted by the source?$
(a) 0.5 x 10^15
(b) 0.5 x 10^17
(c) 5 x 10^17
(d) 5 x 10^15
›Reveal solutionSolution

The number of photons emitted per second is the total power divided by the energy carried by a single photon, E = h nu.

Given: nu = 6.0 x 10^14 Hz, P = 2.0 x 10^-3 W, h = 6.63 x 10^-34 J s.

…

Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. Monochromatic light of frequency 6 x 10^14 Hz is produced by a laser. The power emitted is 2 x 10^-3 W. How many photons per second on an average are emitted by the source? [h = 6.63 x 10^-34 Js]$
(a) 3.98 x 10^19
(b) 1.99 x 10^15
(c) 3 x 10^15
(d) 5 x 10^15
›Reveal solutionSolution

Each photon carries energy E = hν; dividing the total power by this per-photon energy gives the photon emission rate.

E = hν = (6.63 × 10⁻³⁴)(6 × 10¹⁴) = 3.978 × 10⁻¹⁹ J. …

Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level? [h = 6.63 x 10^-34 Js]$
(a) 1.2 x 10^14 Hz
(b) 5.6 x 10^14 Hz
(c) 3.8 x 10^14 Hz
(d) 1.6 x 10^6 Hz
›Reveal solutionSolution

The frequency of emitted radiation during an atomic transition follows Bohr's frequency condition: hν = ΔE.

ΔE = 2.3 eV = 2.3 × 1.6 × 10⁻¹⁹ = 3.68 × 10⁻¹⁹ J. …

Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. The momentum of a photon of light of frequency f is ___.$
(a) hc/f
(b) h/cf
(c) hf/c
(d) hcf
›Reveal solutionSolution

A photon of energy E = hf carries momentum p = E/c.

…

Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. Monochromatic light of frequency 6 x 10^14 Hz is produced by laser. The power emitted is 2 x 10^-3 W. The energy of the photon in this light beam is ___ eV. [h = 6.63 x 10^-34 Js, 1 eV = 1.6 x 10^-19 J]$
(a) 3.0
(b) 3.5
(c) 4.0
(d) 2.5
›Reveal solutionSolution

Photon energy E = h*nu = 3.98 x 10^-19 J; dividing by 1.6 x 10^-19 J/eV gives about 2.5 eV.

Energy of a photon: E = h*nu = (6.63 x 10^-34)(6 x 10^14) = 3.978 x 10^-19 J.

Convert to eV: E = (3.978 x 10^-19)/(1.6 x 10^-19) = 2.49 eV approximately 2.5 eV.

…

Start your 14-day free trial to see the full solution →
GUJCET 2022Set 171 markMCQ
$1.6 \times 10^{-19}$
›Reveal solutionSolution
Photon frequency $f = \Delta E / h$ .

Steps.

$\Delta E = 5.4\text{ eV} = 5.4\times1.6\times10^{-19} = 8.64\times10^{-19}$ J. …
Start your 14-day free trial to see the full solution →
GUJCET 2022Set 171 markMCQ
$Q. What is the shortest wavelength present in the Paschen series of spectral lines? (A) 320 nm (B) 720 nm (C) 840 nm (D) 820 nm$
›Reveal solutionSolution

Shortest wavelength = series limit, $1/\lambda = R_H/9$ .

Concept. The shortest wavelength of a hydrogen series comes from the transition $n=\infty \to n_f$ . For Paschen, $n_f=3$ .

Steps. …
Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
$Q. Monochromatic light of frequency 6x10^14 Hz is produced by laser. Each photon has an energy = ____ J.$
(a) 6x10^14
(b) 4x10^-19
(c) 4x10^-20
(d) 6x10^-14
›Reveal solutionSolution

Photon energy E = hf; substituting f = 6×10¹⁴ Hz gives about 4×10⁻¹⁹ J.

…

Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$Q. Which of the following physical quantity has the dimension of planck constant (h)?$
(a) Angular momentum
(b) Force
(c) Energy
(d) Power
›Reveal solutionSolution

Planck's constant has dimensions of Energy × Time $=[ML^2T^{-2}][T]=[ML^2T^{-1}]$ , which is also the dimension of angular momentum.

…

Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. Energy of photon is E = hf and its momentum is P = h/lambda, where lambda is the wavelength of photon. With this assumption speed of light wave is ___.$
(a) P/E
(b) E/P
(c) EP
(d) (E/P)^2
›Reveal solutionSolution

Speed of light = f x lambda; using f = E/h and lambda = h/P gives speed = E/P.

The speed of a light wave is v = f lambda.

From E = h f: f = E/h.

From P = h/lambda: lambda = h/P.

…

Start your 14-day free trial to see the full solution →

←Previous question11.6 — In an experiment on photoelectric effect, the slope of the cut-off vol…→Next question11.8 — The threshold frequency for a certain metal is 3.3 × 10¹⁴Hz. If light …

Q.A 100 W100\ \text{W}100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm589\ \text{nm}589 nm.

🎓Unlock everything free for 14 days

Q.A $100\ \text{W}$ sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is $589\ \text{nm}$ .

Unlock everything free for 14 days