The multistep biochemical process of photosynthesis involves the formation of gl
ID: 503235 • Letter: T
Question
The multistep biochemical process of photosynthesis involves the formation of glucose (C_6H_32O_2) and O_2 from CO_2 and H_2O. Chlorophyll absorbs light in the region 600 to 700 nm electromagnetic spectrum. What is the minimum number of photo with 680 nm to prepare 1.00 mol of glucose? When light of wavelength 358 nm falls on a surface of potassium metal, the speed of the dislodged electron is 6.40 times 10^5 m/s. a) What is the kinetic energy of the dislodged electron? b) What is the minimum energy needed to dislodge an electron from the surface of the potassium metal? In the photoelectric effect, a photon with energy of 5.3 times 10-19 J strikes an electron in metal. Of this energy 3.6 times 10-19 J is the minimum energy required for the electron to escape from the metal. The remaining energy appears as kinetic energy of the photoelectron. What is the velocity of the photoelectron, assuming it was initially at rest?Explanation / Answer
6) The kinetic energy of the ejected electrons and the energy incident photons are related as:
Ekin = Ep - W
W is the work function, i.e. the minimum energy required to remove an electron from the surface of the metal.
The kinetic energy of the ejected electron is:
a) E_kin = (1/2)mev2
= (1/2) x 9.109×10-31 kg (6.40×105ms¹)2
= (373.10464 / 2 )x 10-21
=1.87 x 10-19 J
The energy of the incident photon is:
E_p = hf = hc/
= 6.626×10-34 Js 3×108 ms¹ / 358×10-9 m
= 5.55 x 10-19 J
b) Hence the minimum energy needed to dislodge one electron is:
W = E_p - E_kin
= 5.55 x 10-19 J - 1.87×1019 J
= 10-19(5.55 - 1.87 ) J
=3.68 X 10-19 J
7) We have the formula as,
1/2mv2 = E'(energy of photon) - E(energy required to remove 'electron')
=> v2 = 2*({5.3-3.6} x 10-19} /9.109 x 10-31
=> v2 = 37.32 x 1010
=> v = 6.109 x 105 meter per sec