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Yanka [14]
3 years ago
12

What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.0×1015Hz?

Chemistry
1 answer:
Anvisha [2.4K]3 years ago
3 0
In order to emit electrons, the cesium will have to absorb photons. Each photon will knock out one electron by transferring its energy to the electron. Therefore, by the principle of energy conservation, the energy of the removed electron will be equal to the energy of the incident photon. That energy is calculated using Planck's equation:

E = hf

E = 6.63 x 10⁻³⁴ * 1 x 10¹⁵
E =  6.63 x 10⁻¹⁹ Joules

The electron will have 6.63 x 10⁻¹⁹ Joules of kinetic energy
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calculate the water potential of a solution of 0.15m sucrose. the solution is at standard temperature.
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Answer:

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

Explanation:

Water potential = Pressure potential + solute potential

P_w=P_p+P_s

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We have :

C = 0.15 M, T = 273.15 K

i = 1

The water potential of a solution of 0.15 m sucrose= P_w

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P_s=-iCRT=-\times 1\times 8.314\times 10^{-2}bar L/mol K\times 273.15 K=-3.406 bar

P_w=0 bar+(-3.406 ) bar

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

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