Question

In a photoelectric effect experiment you illuminate a metal with light of an unknown wavelength and measure the maximum kinetic energy of the photoelectrons to be 0.65 eV. Then you illuminate the same metal with light of a wavelength known to be 2/3 of the first wavelength and measure a maximum kinetic energy of 2.3 eV for the photoelectrons.
(a) Find the first wavelength, in nanometers.
(b) Find the metal's work function, in electron volts.

Answer 1

Answer:

(a) 3.77 x 10^-7 m

(b) 2.65 eV

Explanation:

According to Einstein photoelectric effect equation

$\frac{hc}{\lambda }-W=KE$

where, W is the work function and KE is the kinetic energy and λ be the wavelength.

(a) For first wavelength

$\frac{6.63\times 10^{-34}\times 3\times 10^8}{\lambda }-W=0.65\times 1.6\times 10^{-19}$

$\frac{1.989\times 10^{-25}}{\lambda }-W=1.04 \times 10^{-19}$... (1)

For second wavelength

λ' = 2λ/3

$\frac{6.63\times 10^{-34}\times 3\times 10^8\times 3}{2\times \lambda }-W=2.3\times 1.6\times 10^{-19}$

$\frac{2.984\times 10^{-25}}{\lambda }-W=3.68 \times 10^{-19}$... (2)

Subtract equation (1) from equation (2)

$\frac{0.995\times10^{-25}}{\lambda }=2.64\times 10^{-19}$

λ = 3.77 x 10^-7 m

(b) Put the value of wavelength in equation (1)

$\frac{1.989\times 10^{-25}}{3.77\times10^{-7} }-W=1.04 \times 10^{-19}$

W = 4.24 x 10^-19 J

W = 2.65 eV

Answer 2

Answer:

(a). The first wavelength is 401.0 nm.

(b). The metal's work function is 2.55 eV.

Explanation:

Given that,

Maximum kinetic energy = 0.65 eV

Second wavelength $\lambda_{2}= \dfrac{2}{3}\times\lambda_{1}$

(a). We need to calculate the wavelength

Using equation of work function for first wavelength

$\dfrac{hc}{\lambda_{1}}-W_{0}=0.65\ eV$.....(I)

For second wavelength,

$\dfrac{hc}{\lambda_{2}}-W_{0}=2.3\ eV$

Put the value of second wavelength

$\dfrac{1.5 hc}{\lambda_{1}}-W_{0}=2.3\ eV$....(II)

By subtraction equation (I) from (II)

$0.5\dfrac{hc}{\lambda_{1}}=1.55$

$\lambda_{1}=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{3.1\times1.6\times10^{-19}}$

$\lambda_{1}=4.010\times10^{-7}\ m$

$\lambda_{1}=401.0\times10^{-9}\ m$

$\lambda_{1}=401.0\ nm$

(b). We need to calculate the work function

Using formula of work function

$W_{0}=(\dfrac{hc}{\lambda}-0.55)$

Put the value into the formula

$W_{0}=(\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1.6\times10^{-19}\times401.0\times10^{-9}}-0.55)$

$W_{0}=2.55\ eV$

Hence, (a). The first wavelength is 401.0 nm.

(b). The metal's work function is 2.55 eV.