Panel A shows a ball shortly after being thrown upward. Panel B shows the same ball in an instant on its way down. Suppose air r
esistance can be ignored. Which statement is true? a) Both potential and kinetic energy are higher in Panel A than in Panel B. b) Both the potential and kinetic energy are higher in Panel B than in Panel A. c) The potential energy is higher in Panel A, but the kinetic energy is higher in Panel B. d) The potential energy is higher in Panel B, but the kinetic energy is higher in Panel A.
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This question is incomplete, the complete question is;
When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.28 × 10⁻¹⁹ J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.
Answer:
the required wavelength of light is 161.9 nm
Explanation:
Given the data in the question;
Let us represent work function of the metal by W.
Now, using Einstein photoelectric effect equation;
= W +
hc/λ = W + ------- let this be equation 1
we solve for W
W = hc/λ -
given that; λ = 221 nm = 2.21 × 10⁻⁷ m, = 3.28 × 10⁻¹⁹ J
we know that speed of light c = 3 × 10⁸ m/s and Planck's constant h = 6.626 × 10⁻³⁴ Js.
so we substitute
W = [( (6.626 × 10⁻³⁴)(3 × 10⁸) )/2.21 × 10⁻⁷ ] - 3.28 × 10⁻¹⁹
W = 8.99457 × 10⁻¹⁹ - 3.28 × 10⁻¹⁹
W = 5.71457 × 10⁻¹⁹ J
Now, to determine λ for which maximum kinetic energy is double
so;
= double = 2( 3.28 × 10⁻¹⁹ J ) = 6.56 × 10⁻¹⁹ J
from from equation 1
we solve for λ'
λ' = hc / W +
we substitute
λ' = ( (6.626 × 10⁻³⁴)(3 × 10⁸) ) / ( (5.71457 × 10⁻¹⁹ J) + ( 6.56 × 10⁻¹⁹ J ))
λ' = 1.9878 × 10⁻²⁴ / 1.227457 × 10⁻¹⁸
λ' = 1.619 × 10⁻⁷ m
λ' = 161.9 nm
Therefore, the required wavelength of light is 161.9 nm