Question

If wavelength of $470\text{nm}$ falls on the surface of potassium metals, eletcrons are emitted with a velocity of $6.4\times {10}^5{\text{m s}}^{-1}$.
The work function of potassium metal is:

• A. $2.36\times {10}^{-19}\text{J}$
• B. $1.86\times {10}^{-21}\text{J}$
• C. $4.25\times {10}^{-30}\text{J}$
• D. $4.2\times {10}^{-18}\text{J}$

Answer 1

The correct option is A $2.36\times {10}^{-19}\text{J}$

Given: $\text{Wavelength}\left(\lambda \right)=470\text{nm}$
$\text{Velocity}\left(v\right)=6.4\times {10}^4{\text{m s}}^{-1}$
We know,
$K.E.=E-\varphi$
Where, $\varphi =\text{Work function of the metal}$
$\therefore \varphi =E-K.E.$

$E=\frac{hc}{\lambda }=\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{470\times {10}^{-9}}$

$=4.22\times {10}^{-19}\text{J}$

$K.E.=\frac{1}{2}m{\text{v}}^2=\frac{1}{2}\times 9.1\times {10}^{-31}\times (6.4\times {10}^5{)}^2$

$=1.86\times {10}^{-19}\text{J}$

Therefore, $\varphi =4.22\times {10}^{-19}-1.86\times {10}^{-19}$
$=2.36\times {10}^{-19}\text{J}$