Question

The six metals have the work functions, W.

Part A 

Rank these metals on the basis of their cutoff frequency. Rank from largest to smallest. To rank items as equivalent, overlap them.

Part B 

Rank these metals on the basis of the maximum wavelength of light needed to free electrons from their surface. Rank from largest to smallest. To rank items as equivalent, overlap them.

Part C 

Each metal is illuminated with 400 nm (3.10 eV) light. Rank the metals on the basis of the maximum kinetic energy of the emitted electrons. (If no electrons are emitted from a metal, the maximum kinetic energy is zero, so rank that metal as smallest.) Rank from largest to smallest. To rank items as equivalent, overlap them.

Cesium= w= 2.1 eV Aliminium= w= 4.1 eV Beryllium= 5.0 eV Potassium= 2.3 eV Platinium= w= 6.4 eV Magnisium=w= 3.7 eV

Answer 1

Concepts and reason

The concept required to solve the given problem is work function and cut off frequency.

Initially, calculate the work function of different materials based on their cut off frequency. Later on, rank the metals based on the wavelength of light needed to free electrons from their surface and finally rank the metals on the basis of the maximum kinetic energy of the emitted electrons will be calculated.

Fundamentals

Cut off frequency is the minimum frequency required to remove electrons from the metal surface.

The expression for the Einstein photoelectric equation is,

$K = hv - {\phi _0}$

Here, K is the kinetic energy of the particle, h is the Planck’s constant, $v$ is the frequency of incident light, and $\phi$ is the work function.

(A)

The relationship between work function and cut off frequency is,

$\phi = hv$

Here, $\phi$ is the work function, $h$ is the Planck’s constant, and $v$ is the frequency of incident radiation.

From the above equation, the work function is directly proportional to the cut-off frequency and when a metal has low a work function, it is easy for the light of low frequency to emit electrons. Therefore, low work function results in low cut-off frequency.

(B)

The work function and wavelength are related as,

$\phi = h\frac{c}{\lambda }$

Here, $\phi$ is the work function, c is the speed of light, h is the planck’s constant, and $\lambda$ is the wavelength of light.

From the equation, the work function is inversely related with the wavelength, so the metals with smallest work function will have maximum wavelength.

(C)

The Einstein photoelectric equation is,

$K = hv - \phi$

Here, K is the kinetic energy of the particle, h is the Planck’s constant, $v$ is the frequency of incident light, and $\phi$ is the work function.

For cesium:

Substitute 3.10 eV for $hv$ and 2.1 eV for $\phi$ in the above equation $K = hv - \phi$.

$\begin{array}{c}\\K\left( {{\rm{Cesium}}} \right) = hv - \phi \\\\ = 3.10{\rm{ eV}} - {\rm{2}}{\rm{.1 eV}}\\\\{\rm{ = 1 eV}}\\\\{\rm{ = 1}}{\rm{.6}} \times {\rm{1}}{{\rm{0}}^{ - 19}}{\rm{ J}}\\\end{array}$

For potassium:

Substitute 3.10 eV for $hv$ and 2.30 eV for $\phi$ in the above equation $K = hv - \phi$.

$\begin{array}{c}\\K\left( {{\rm{potassium}}} \right) = hv - \phi \\\\ = 3.10{\rm{ eV}} - {\rm{2}}{\rm{.30 eV}}\\\\{\rm{ = 0}}{\rm{.80 eV}}\\\\{\rm{ = 1}}{\rm{.28}} \times {\rm{1}}{{\rm{0}}^{ - 19}}{\rm{ J}}\\\end{array}$

For magnesium:

Substitute 3.10 eV for $hv$ and 3.70 eV for $\phi$ in the above equation $K = hv - \phi$.

$\begin{array}{c}\\K\left( {{\rm{magnesium}}} \right) = hv - \phi \\\\ = 3.10{\rm{ eV}} - {\rm{3}}{\rm{.70 eV}}\\\\{\rm{ = }} - {\rm{0}}{\rm{.60 eV}}\\\\{\rm{ = }} - {\rm{0}}{\rm{.96}} \times {\rm{1}}{{\rm{0}}^{ - 19}}{\rm{ J}}\\\end{array}$

Similarly, for other materials, the energy will be negative and hence no ejection of electrons will take places from its surfaces.

Ans: Part A

Rank these metals on the basis of their cut off frequency from largest to smallest is Pt>Be>Al/Mg>K>Cs.

Part B

Rank these metals on the basis of the maximum wavelength of light needed to free electrons from their surface from largest to smallest is ${\rm{Cs}} > {\rm{K}} > {\rm{Mg}} > {\rm{Al}} > {\rm{Be}} > {\rm{Pt}}$.

Part C

Rank the metals on the basis of their maximum kinetic energy of the emitted electrons from largest to smallest is ${\rm{Cs > K > Mg = Al = Be = Pt}}$.