Question

Determine the region of the electron spectrum which liberates photo-electrons from potassium given that the function of potassium is $2.24eV$.
A.$< 5520A$
B.$> 5520A$
C.$> 8.17A$
D.$< 8.17A$

Answer 1

The work function is given in the question, with the help of work function and Planck’s constant we can calculate the frequency and then the frequency can be substituted in the wavelength formula gives the region of the electron spectrum which liberate photo-electrons from potassium.

Complete answer:
Given the work function of potassium is $2.24eV$.
This work function is in electron volts; it should be converted into Coulombs.
Thus, this value should be multiplied with $1.6 \times {10^{ - 19}}columbs$
The planck’s constant h has the value of $6.624 \times {10^{ - 34}}{m^2}kg{\left( s \right)^{ - 1}}$
The frequency can be calculated from planck’s constant and work function has to be substituted in the frequency
$\nu = \dfrac{\omega }{h} = \dfrac{{2.24 \times 1.6 \times {{10}^{ - 19}}}}{{6.6 \times {{10}^{ - 34}}}}Hz = 5.43 \times {10^{14}}Hz$
Substitute the value of frequency in the wavelength. The velocity of light can be $3 \times {10^8}cm$
The frequency calculated is $5.43 \times {10^{14}}Hz$
Thus, wavelength will be equal to $\lambda = \dfrac{c}{\vartheta }$
Substitute the velocity of light and frequency in wavelength, we will get
$\lambda = \dfrac{{3 \times {{10}^8}}}{{5.43 \times {{10}^{14}}}} = 5.52 \times {10^{ - 7}}m$
The obtained wavelength will be $5.52 \times {10^{ - 7}}m$
But angstroms are the units used to express the wavelength
$1Angstorm = {10^{ - 8}}cm$
The obtained wavelength is $5.52 \times {10^{ - 7}}m$, after converting this wavelength to angstroms we get $5520{A^0}$
Thus, the wavelength shorter than $5520{A^0}$ is the region of the electron spectrum which liberates photo-electrons from potassium given that the function of potassium is $2.24eV$.
Thus, option A is the correct one.

Note:
The angstroms are also the units used to express the work function that can be expressed in coulombs and electron volts. Thus, one must be careful in converting the units from coulombs to electron volts and the frequency has the units of hertz.