Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) thres

Exercises · 2.52

Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and (b) Planck's constant.

$\lambda$ (nm)	500	450	400
$v \times 10^{-5}$ (cm s $^{-1}$ )	2.55	4.35	5.35

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Concept understanding — Photon Energy Calculation

Photon Energy Calculation

The Intuition First

Imagine you're holding a rope tied to a wall. If you flick your wrist once, a single pulse travels down the rope. If you flick faster — more frequently — each pulse carries more energy; the rope vibrates more violently. Light behaves the same way. A photon is the smallest possible "flick" of the electromagnetic field — a single, indivisible packet of light energy.

What determines how much energy that one photon carries? Two things: how fast the wave is oscillating (its frequency), and a universal constant that connects the wave world to the particle world.

The Precise Statement

The energy $E$ of a single photon is directly proportional to its frequency $f$ (or inversely proportional to its wavelength $\lambda$ ). The proportionality constant is Planck's constant $h$ , one of the most fundamental numbers in physics.

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

Where:

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

Why Two Forms?

The first form $E = hf$ is the most direct: higher frequency means higher energy. The second form $E = hc/\lambda$ is often more practical because wavelength is easier to measure than frequency. Since $c = f\lambda$ , you can always swap between them.

What This Tells You

Blue light (short wavelength, high frequency) has more energy per photon than red light (long wavelength, low frequency).
Gamma rays have enormous photon energies; radio waves have tiny photon energies.
The energy is quantised — you cannot have half a photon. Either the full energy $hf$ is absorbed/emitted, or none at all.

A Worked Example

Question: Calculate the energy of a single photon of violet light with wavelength $400 \ \text{nm}$ .

Step 1: Convert wavelength to metres.

400 \ \text{nm} = 400 \times 10^{-9} \ \text{m} = 4.00 \times 10^{-7} \ \text{m}

Step 2: Use $E = hc/\lambda$ .

E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{4.00 \times 10^{-7}}

Step 3: Compute.

E = \frac{1.9878 \times 10^{-25}}{4.00 \times 10^{-7}} = 4.97 \times 10^{-19} \ \text{J}

Note

This is an incredibly tiny amount of energy — about $5 \times 10^{-19}$ joules. That's why we often use electronvolts (eV) for photon energies in atomic physics. $1 \ \text{eV} = 1.602 \times 10^{-19} \ \text{J}$ , so this photon has about $3.1 \ \text{eV}$ .

Common Mistake to Avoid …

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Convert each wavelength to a frequency, each speed to a kinetic energy, and fit Einstein's photoelectric equation $\tfrac{1}{2}m_e v^2 = h\nu - W_0$ through the three data points. The slope gives $h \approx 6.7 \times 10^{-34}$ J s and the intercept gives $W_0 \approx 3.7 \times 10^{-19}$ J, i.e. a threshold wavelength $\boxed{\lambda_0 \approx 540\ \text{nm}}$ .

Setting Up: What the Data Mean

Sodium is irradiated at three wavelengths, and the ejected photoelectrons' speeds are measured:

$\lambda$ (nm)	500	450	400
$v$ ( $10^{5}$ m s $^{-1}$ )	2.55	4.35	5.35

A note on units, stated honestly: the book's column header reads " $v \times 10^{-5}$ (cm s $^{-1}$ )". Interpreted literally as cm s $^{-1}$ the kinetic energies would come out around $10^{-24}$ J — about 100,000 times smaller than the photon energies ( $\sim 4 \times 10^{-19}$ J), which would make every kinetic energy negligible and the data unusable. Physically consistent numbers (and the standard treatment of this exercise) require $v$ in units of $10^{5}$ m s $^{-1}$ , which is what we use below.

Step 1 — Frequencies of the Three Radiations

\nu = \frac{c}{\lambda}: \quad \nu_1 = \frac{3.0 \times 10^{8}}{500 \times 10^{-9}} = 6.00 \times 10^{14}\ \text{Hz}, \quad \nu_2 = 6.67 \times 10^{14}\ \text{Hz}, \quad \nu_3 = 7.50 \times 10^{14}\ \text{Hz}

Step 2 — Kinetic Energies of the Photoelectrons

With $m_e = 9.11 \times 10^{-31}$ kg and $K = \tfrac{1}{2}m_e v^2$ :

K_1 = \tfrac{1}{2}(9.11 \times 10^{-31})(2.55 \times 10^{5})^2 = 2.96 \times 10^{-20}\ \text{J}

K_2 = \tfrac{1}{2}(9.11 \times 10^{-31})(4.35 \times 10^{5})^2 = 8.62 \times 10^{-20}\ \text{J}

K_3 = \tfrac{1}{2}(9.11 \times 10^{-31})(5.35 \times 10^{5})^2 = 1.30 \times 10^{-19}\ \text{J}

Step 3 — Planck's Constant from the Slope

Einstein's photoelectric equation, $K = h\nu - W_0$ , is a straight line in $K$ vs $\nu$ with slope $h$ . Using the two extreme points: …

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Setting Up: What the Data Mean

Step 1 — Frequencies of the Three Radiations

Step 2 — Kinetic Energies of the Photoelectrons

Step 3 — Planck's Constant from the Slope

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