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Q.Photoelectrons are emitted from a metal surface when illuminated with UV light of wavelength 330 nm. The minimum amount of energy required to emit the electrons from the surface is 3.5×10−19 J3.5 \times 10^{-19}\ \text{J}. Calculate:

(i) the energy of the incident radiation, and
(ii) the kinetic energy of the photoelectron.
CBSECBSE Class XII Board 2022Subjective· 3mImportance★★★★★
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The photoelectric effect equation Ephoton=ϕ+KmaxE_{\text{photon}} = \phi + K_{\text{max}} links the incident photon energy, work function, and maximum kinetic energy. For 330 nm UV light, the photon energy is 6.0×10−19 J6.0 \times 10^{-19}\ \text{J}, and the emitted electron's kinetic energy is 2.5×10−19 J2.5 \times 10^{-19}\ \text{J}.

The Concept: Why This Works

The photoelectric effect is one of those rare places in physics where a single equation tells you everything. When light hits a metal surface, each photon can transfer its entire energy to a single electron. But the electron can't just fly out — it first needs to overcome the "work function" ϕ\phi, which is the minimum energy binding it to the metal. Whatever energy the photon has left over becomes the electron's kinetic energy.

Think of it like this: the photon hands the electron a certain amount of money (energy). The electron must first pay a tax (the work function) to leave the metal. Whatever remains is its spending money (kinetic energy). The equation is:

Ephoton=ϕ+KmaxE_{\text{photon}} = \phi + K_{\text{max}}

Where Ephoton=hcλE_{\text{photon}} = \frac{hc}{\lambda}, ϕ\phi is the work function, and KmaxK_{\text{max}} is the maximum kinetic energy of the emitted photoelectron.

Step-by-Step Solution

1. Find the energy of the incident photon.

The photon's energy depends only on its wavelength (or frequency). For light of wavelength λ=330 nm=330×10−9 m\lambda = 330\ \text{nm} = 330 \times 10^{-9}\ \text{m}, we use:

E=hcλE = \frac{hc}{\lambda}

Where Planck's constant h=6.63×10−34 J⋅sh = 6.63 \times 10^{-34}\ \text{J·s} and the speed of light c=3.0×108 m/sc = 3.0 \times 10^{8}\ \text{m/s}.

E=(6.63×10−34)(3.0×108)330×10−9E = \frac{(6.63 \times 10^{-34})(3.0 \times 10^{8})}{330 \times 10^{-9}}

Let's compute step by step. First, the numerator:

hc=6.63×10−34×3.0×108=1.989×10−25 J⋅mhc = 6.63 \times 10^{-34} \times 3.0 \times 10^{8} = 1.989 \times 10^{-25}\ \text{J·m}

Now divide by the wavelength:

E=1.989×10−253.30×10−7=6.027×10−19 JE = \frac{1.989 \times 10^{-25}}{3.30 \times 10^{-7}} = 6.027 \times 10^{-19}\ \text{J}

Rounding to two significant figures (matching the given data), the incident photon energy is:

Ephoton=6.0×10−19 JE_{\text{photon}} = 6.0 \times 10^{-19}\ \text{J}

Tip

A quick check: for UV light around 300–400 nm, photon energies are typically in the 10−1910^{-19} J range. If you ever get an answer wildly different (like 10−1710^{-17} or 10−2110^{-21} J), you've likely misplaced a decimal in the wavelength conversion.

2. Calculate the kinetic energy of the photoelectron.

The problem gives the work function directly: ϕ=3.5×10−19 J\phi = 3.5 \times 10^{-19}\ \text{J}. This is the "minimum amount of energy required to emit the electrons" — that's exactly the definition of the work function.

Now apply the photoelectric equation:

Kmax=Ephoton−ϕK_{\text{max}} = E_{\text{photon}} - \phi

Kmax=(6.0×10−19)−(3.5×10−19)K_{\text{max}} = (6.0 \times 10^{-19}) - (3.5 \times 10^{-19})

Kmax=2.5×10−19 JK_{\text{max}} = 2.5 \times 10^{-19}\ \text{J} …

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