Q.Monochromatic light of frequency $6.0 \times 10^{14}\ \text{Hz}$ is produced by a laser. The power emitted is $2.0 \times 10^{-3}\ \text{W}$.

Concept: Photon Energy — each photon carries a discrete quantum of energy $E = h\nu$, where $h$ is Planck’s constant and $\nu$ is the frequency.

(a) Energy of one photon:

(b) Power $P$ is energy per second. If $n$ photons are emitted each second, total energy per second is $nE = P$. Hence:

Why Photon Energy Matters Here

Light is not a continuous stream of energy — it comes in discrete packets called photons. Each photon carries a specific energy that depends only on the frequency (or wavelength) of the light, not on the intensity. The laser's power tells us how much total energy is delivered per second. To find how many photons leave the laser each second, we simply divide the total energy per second (power) by the energy carried by one photon.

This is a clean, two-step problem: first find the energy of one photon, then count how many such photons make up the total power.

Step-by-Step Solution

1. Energy of a single photon

The energy $E$ of one photon is given by the Planck-Einstein relation:

where

$h = 6.626 \times 10^{-34}\ \text{J·s}$ (Planck's constant)

$\nu = 6.0 \times 10^{14}\ \text{Hz}$ (frequency)

Substitute:

Rounding to two significant figures (matching the given data):

2. Number of photons emitted per second

Power $P$ is energy per unit time. If each photon carries energy $E$, then the number of photons emitted per second $n$ satisfies:

So:

Given $P = 2.0 \times 10^{-3}\ \text{W}$ (which is $2.0 \times 10^{-3}\ \text{J/s}$):

Rounding to two significant figures:

Method: Photon Energy Approach (using Planck's relation)

This problem uses the fundamental idea that light energy comes in discrete packets called photons. The energy of each photon depends only on the frequency of the light, not on the power. Power tells us how much total energy is delivered per second, so dividing that by the energy per photon gives the number of photons per second.

(a) Energy of a single photon

Step 1: Recall Planck's relation — the energy of one photon is directly proportional to its frequency:

where $h = 6.63 \times 10^{-34}\ \text{J·s}$ (Planck's constant) and $f$ is the frequency in hertz.

Step 2: Substitute the given frequency:

Step 3: Multiply the numbers and the powers of ten separately:

So $E = 39.78 \times 10^{-20}\ \text{J} = 3.978 \times 10^{-19}\ \text{J}$

Step 4: Round to two significant figures (since the given frequency has two significant figures):

(b) Number of photons emitted per second

Step 1: Understand what power means. Power $P = 2.0 \times 10^{-3}\ \text{W}$ means the source delivers $2.0 \times 10^{-3}$ joules of energy each second.

Step 2: If each photon carries $E$ joules, then the number of photons emitted per second, $n$, is:

Step 3: Substitute the values:

Step 4: Divide the coefficients and subtract the exponents:

So $n = 0.5 \times 10^{16} = 5.0 \times 10^{15}$

Final answers:

1. Energy of one photon = $4.0 \times 10^{-19}\ \text{J}$
2. Number of photons emitted per second = $5.0 \times 10^{15}$

Common Mistakes & How to Avoid Them

Mistake 1: Using the wrong formula for photon energy

Students often confuse $E = hf$ with $E = \frac{hc}{\lambda}$. Both are correct, but the first is direct when frequency is given. The second requires an extra step (converting frequency to wavelength) and introduces more places for error — like using the wrong value for $c$ or forgetting to convert units.

How to avoid: When frequency is given, use $E = hf$ directly. Only switch to $E = hc/\lambda$ when wavelength is provided. Memorise both forms but pick the one that matches the data.

Mistake 2: Forgetting the value of Planck's constant or using the wrong one

Planck's constant $h$ is $6.63 \times 10^{-34}\ \text{J·s}$. Some students use $6.6 \times 10^{-34}$ (acceptable in some boards, but risky) or accidentally use $h = 4.14 \times 10^{-15}\ \text{eV·s}$ when the answer is expected in joules.

How to avoid: Write down $h = 6.63 \times 10^{-34}\ \text{J·s}$ at the top of your working. If the question asks for energy in joules (which it usually does unless specified), stick to this value. If you must use the eV version, convert at the end — don't mix units mid-calculation.

Mistake 3: Incorrect exponent handling in part (a)

The calculation is:

Students often add exponents incorrectly: $10^{-34} \times 10^{14} = 10^{-20}$, not $10^{-48}$ or $10^{-20}$ with a sign error. Also, they sometimes forget to multiply the coefficients: $6.63 \times 6.0 \approx 39.78$, not $3.978$.

How to avoid: Separate the calculation into two parts:

• Multiply the coefficients: $6.63 \times 6.0 = 39.78$
• Add the exponents: $-34 + 14 = -20$
• Combine: $39.78 \times 10^{-20} = 3.978 \times 10^{-19}$

Then round to appropriate significant figures (here, two significant figures from the given data gives $4.0 \times 10^{-19}\ \text{J}$).

Mistake 4: Confusing power with energy in part (b)

Power $P = 2.0 \times 10^{-3}\ \text{W}$ means $2.0 \times 10^{-3}\ \text{J}$ of energy is emitted per second. Some students treat power as the total energy or forget that it's already a rate.

How to avoid: Write down what power means: $P = \frac{E_{\text{total}}}{t}$. For $t = 1\ \text{s}$, $E_{\text{total}} = P \times 1 = P$. So the number of photons per second is:

Mistake 5: Dividing in the wrong order

Students sometimes compute $E/P$ instead of $P/E$, getting a tiny fraction instead of a large number.

How to avoid: Check the units. You want photons per second, which has units of $\text{s}^{-1}$. $P$ has units $\text{J/s}$, $E$ has units $\text{J}$. So $P/E$ gives $\frac{\text{J/s}}{\text{J}} = \text{s}^{-1}$, which is correct. $E/P$ gives seconds — a time, not a rate.

Mistake 6: Arithmetic errors in part (b)

Common errors: dividing coefficients as $2.0/4.0 = 0.5$ but then writing $0.5 \times 10^{-16}$ (sign error on exponent), or forgetting that $10^{-3}/10^{-19} = 10^{16}$.

How to avoid: Again, separate coefficient and exponent:

• Coefficients: $2.0 / 4.0 = 0.5$
• Exponents: $-3 - (-19) = -3 + 19 = 16$
• Combine: $0.5 \times 10^{16} = 5.0 \times 10^{15}$

Final Answers