Question

The threshold wavelength for a metal surface whose photoelectric work function is 3.313 eV is​

Answer 1

Given:

Photoelectric work function is 3.313 eV

To Find:

Threshold wavelength for a metal surface

Solution:

Given that,

$\phi$ = 3.313 eV

Converting into Joules

$\phi$ = 3.313 x 1.6 x 10⁻¹⁹ J

$\phi$ = 5.30 x 10⁻¹⁹ J

We know that,

$\red{\bigstar \ \boxed{\rm \lambda_{0}=\dfrac{hc}{\phi}}}$

Here,

• $\lambda_{0}$ = threshold wavelength

• h = planck's constant

• c = velocity of light

• $\phi$ = work function

We know that,

• h = 6.63 x 10⁻³⁴

• c = 3 x 10⁸

We found out that,

• $\phi$ = 5.30 x 10⁻¹⁹ J

Substituting the values,

We get,

$\blue{\rm \implies \lambda_{0}=\dfrac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{5.30 \times 10^{-19}}}$

$\green{\rm \implies \lambda_{0}=\dfrac{19.89 \times 10^{-26}}{5.30 \times 10^{-19}}}$

$\orange{\implies \rm \lambda_{0}=3.752 \times 10^{-7}}$

$\pink{\implies \rm \lambda_{0}=3.752 \ A^{\circ}}$

Hence,

Threshold wavelength = 3752 A°