
The photoelectric work function for a metal is $6 \times \mathop {10}\nolimits^{ - 19} $J. Threshold wavelength is equal to
A. 2305A
B. 3315A
C. 3625A
D. 4575A
Answer
586.2k+ views
HINT: Threshold wavelength is the maximum wavelength that the incident radiation must have in order to execute the photoelectric effect. Firstly, we need to have some basic knowledge about the dual nature of matter to understand this question. The formula being used in this question is Einstein’s Photoelectric equation ($\phi \circ = \dfrac{{\mathop h\nolimits_c }}{{\lambda \circ }}$) and the relation between maximum kinetic energy and stopping potential.
Complete step by step solution:
According to the given question, firstly, we will talk about the photoelectric effect in which the concept of threshold wavelength comes in my mind.
Photoelectric effect is that effect in which emission of electrons when electromagnetic radiation such as light hits onto a material.
To execute a photoelectric effect, the wavelength of the incident radiation should be lesser than the threshold wavelength of the surface. In most cases which is metal.
Let us look at the formula of the threshold wavelength
$\phi = h \times v \circ $
Or
$\phi = \dfrac{{hc}}{{\lambda \circ }}$……. (1)
Where $v \circ = \dfrac{c}{{\lambda \circ }}$
$\phi $= it is the work function of the surface
H= Planck’s constant=$6.63 \times \left( {\mathop {10}\nolimits^{ - 34} } \right)$Js
$v \circ $= speed of light
$\lambda \circ $= threshold wavelength of photoelectric effect
Hence, from the equation (1), we get
Work function$\phi = \dfrac{{hc}}{\lambda }$
$ \Rightarrow \lambda = \dfrac{{hc}}{\begin{gathered}
\phi \\
\\
\end{gathered} }$
$ \Rightarrow \dfrac{{6.626 \times \mathop {10}\nolimits^{ - 34} \times 3 \times \mathop {10}\nolimits^8 }}{{6 \times \mathop {10}\nolimits^{ - 19} }} = 3315A$
NOTE: Threshold wavelength should not be confused with the threshold frequency. It is the minimum frequency of incident radiation below which the photoelectric effect is not possible. Work function is defined as the amount of work that is needed to be done to remove an electron from a solid.
Complete step by step solution:
According to the given question, firstly, we will talk about the photoelectric effect in which the concept of threshold wavelength comes in my mind.
Photoelectric effect is that effect in which emission of electrons when electromagnetic radiation such as light hits onto a material.
To execute a photoelectric effect, the wavelength of the incident radiation should be lesser than the threshold wavelength of the surface. In most cases which is metal.
Let us look at the formula of the threshold wavelength
$\phi = h \times v \circ $
Or
$\phi = \dfrac{{hc}}{{\lambda \circ }}$……. (1)
Where $v \circ = \dfrac{c}{{\lambda \circ }}$
$\phi $= it is the work function of the surface
H= Planck’s constant=$6.63 \times \left( {\mathop {10}\nolimits^{ - 34} } \right)$Js
$v \circ $= speed of light
$\lambda \circ $= threshold wavelength of photoelectric effect
Hence, from the equation (1), we get
Work function$\phi = \dfrac{{hc}}{\lambda }$
$ \Rightarrow \lambda = \dfrac{{hc}}{\begin{gathered}
\phi \\
\\
\end{gathered} }$
$ \Rightarrow \dfrac{{6.626 \times \mathop {10}\nolimits^{ - 34} \times 3 \times \mathop {10}\nolimits^8 }}{{6 \times \mathop {10}\nolimits^{ - 19} }} = 3315A$
NOTE: Threshold wavelength should not be confused with the threshold frequency. It is the minimum frequency of incident radiation below which the photoelectric effect is not possible. Work function is defined as the amount of work that is needed to be done to remove an electron from a solid.
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