For a metal of work function 6.6 eV, which of the following wavelengths of incident radia tion does not give rise to the photoelectric effect? (Take Planck’s constant as 6.6 × 10³ Js)

First, write down the relation between the energy of a photon and its wavelength:

$E\; =\; \backslash frac\{h\; c\}\{\backslash lambda\}$

where

• $h\; =\; 6.6\; \backslash times\; 10^\{-34\}\backslash ;\; \backslash text\{J·s\}$ (Planck’s constant – we use the standard value even though the statement gives a typo),
• $c\; =\; 3.0\; \backslash times\; 10^\{8\}\backslash ;\; \backslash text\{m/s\}$ (speed of light),
• $\backslash lambda$ is the wavelength of the incident radiation.

Convert each wavelength from nanometres to metres ($1\backslash ;\backslash text\{nm\}=10^\{-9\}\backslash ;\backslash text\{m\}$) and calculate the corresponding photon energy.

1. λ = 50 nm

$\backslash lambda\; =\; 50\; \backslash times\; 10^\{-9\}\backslash ;\; \backslash text\{m\}=5.0\; \backslash times\; 10^\{-8\}\backslash ;\; \backslash text\{m\}$

$E\; =\; \backslash frac\{6.6\; \backslash times\; 10^\{-34\}\backslash times\; 3.0\; \backslash times\; 10^\{8\}\}\{5.0\; \backslash times\; 10^\{-8\}\}\; =\; \backslash frac\{1.98\; \backslash times\; 10^\{-25\}\}\{5.0\; \backslash times\; 10^\{-8\}\}\; =\; 3.96\; \backslash times\; 10^\{-18\}\backslash ;\; \backslash text\{J\}$

Convert to electron‑volts ($1\backslash ;\backslash text\{eV\}=1.60\; \backslash times\; 10^\{-19\}\backslash ;\; \backslash text\{J\}$):

$E\; =\; \backslash frac\{3.96\; \backslash times\; 10^\{-18\}\}\{1.60\; \backslash times\; 10^\{-19\}\}\; \backslash approx\; 24.8\backslash ;\; \backslash text\{eV\}$

Since 24.8 eV > 6.6 eV, the photoelectric effect occurs.

2. λ = 100 nm

$\backslash lambda\; =\; 1.0\; \backslash times\; 10^\{-7\}\backslash ;\; \backslash text\{m\}$

$E\; =\; \backslash frac\{6.6\; \backslash times\; 10^\{-34\}\backslash times\; 3.0\; \backslash times\; 10^\{8\}\}\{1.0\; \backslash times\; 10^\{-7\}\}\; =\; 1.98\; \backslash times\; 10^\{-18\}\backslash ;\; \backslash text\{J\}$

$E\; =\; \backslash frac\{1.98\; \backslash times\; 10^\{-18\}\}\{1.60\; \backslash times\; 10^\{-19\}\}\; \backslash approx\; 12.4\backslash ;\; \backslash text\{eV\}$

Again, 12.4 eV > 6.6 eV ��� photoelectric effect.

3. λ = 150 nm

$\backslash lambda\; =\; 1.5\; \backslash times\; 10^\{-7\}\backslash ;\; \backslash text\{m\}$

$E\; =\; \backslash frac\{6.6\; \backslash times\; 10^\{-34\}\backslash times\; 3.0\; \backslash times\; 10^\{8\}\}\{1.5\; \backslash times\; 10^\{-7\}\}\; =\; 1.32\; \backslash times\; 10^\{-18\}\backslash ;\; \backslash text\{J\}$

$E\; =\; \backslash frac\{1.32\; \backslash times\; 10^\{-18\}\}\{1.60\; \backslash times\; 10^\{-19\}\}\; \backslash approx\; 8.3\backslash ;\; \backslash text\{eV\}$

8.3 eV > 6.6 eV → photoelectric effect.

4. λ = 200 nm

$\backslash lambda\; =\; 2.0\; \backslash times\; 10^\{-7\}\backslash ;\; \backslash text\{m\}$

$E\; =\; \backslash frac\{6.6\; \backslash times\; 10^\{-34\}\backslash times\; 3.0\; \backslash times\; 10^\{8\}\}\{2.0\; \backslash times\; 10^\{-7\}\}\; =\; 9.9\; \backslash times\; 10^\{-19\}\backslash ;\; \backslash text\{J\}$

$E\; =\; \backslash frac\{9.9\; \backslash times\; 10^\{-19\}\}\{1.60\; \backslash times\; 10^\{-19\}\}\; \backslash approx\; 6.2\backslash ;\; \backslash text\{eV\}$

Here, 6.2 eV < 6.6 eV, which is **below the work function** of the metal. Therefore, photons of 200 nm wavelength do **not** have enough energy to liberate electrons, and the photoelectric effect does not occur.

Hence, the wavelength that does not give rise to the photoelectric effect is **200 nm**.