Atomic StructureHard
Question
The threshold wavelength for ejection of electrons from a metal is 330 nm. The work function for the photoelectric emission from the metal is (h = 6.6 × 10–34 J-s)
Options
A.1.2 × 10–18 J
B.6.0 × 10–19 J
C.1.2 × 10–20 J
D.6.0 × 10–12 J
Solution
$\phi = \frac{hc}{\lambda} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{330 \times 10^{- 9}} = 6 \times 10^{- 19}\text{ J}$
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
The wavelength of spectral line obtained in the spectrum of ${LI}^{2 +}$ ion, when the transition takes place between tw...Consider the ground state of Cr atom (Z = 24). The number of electrons with the azimuthal quantum numbers I = 1 and 2 ar...Bohr’s model may be applied to...The time period of revolution in the third orbit of Li2+ ion is x second. The time period of revolution in the second or...Beryllium fourth electron will have the four quantum numbers :-...