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trasher [3.6K]
1 year ago
8

An 80 ev electron impinges upon a potential barrier 100 ev high and 0. 2 nm thick. what is the probability the electron will tun

nel through the barrier?
Chemistry
1 answer:
AleksAgata [21]1 year ago
5 0

In the above question the probability the electron will tunnel through the barrier is 0.011%.

<h3>What do you mean by electron volt?</h3>

An electron volt is the amount of energy required to move a charge equal to 1e⁻ across a potential difference of 1eV.

To calculate the probability the electron will tunnel through the barrier is calculated as -

Energy of electron

E=80eV

=80×1.6×10⁻¹⁹

=128×10⁻¹⁹

Height of the barrier U=100 eV

=100×1.6×10⁻¹⁹J

Thickness L=0.2×10⁻⁹m

Probability T=e⁻²cl

C=√2m(U-E)/h

=√[2×1.67×10⁻²⁷(160-128)×10⁻¹⁹] / 1.055×10⁻³⁴

=10.34×10⁻²³/1.055×10⁻³⁴

=9.8×10¹¹

2CL=2×9.8×10¹¹×0.2×10⁻⁹

=3.92×10²

T=e⁻²cl

= 1/e⁻³⁹²/¹⁰⁰⁰⁰

=0.0198%

=0.011%

Hence ,the probability the electron will tunnel through the barrier is 0.011%.

Learn more about electron volt ,here:

brainly.com/question/17136195

#SPJ4

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When 500.0 g of water is decomposed by electrolysis and the yield of hydrogen is only 75.3%, how much hydrogen chloride can be m
Evgen [1.6K]

The amount of hydrogen chloride that can be made is 1064 g

Why?

The two reactions are:

2H₂O → 2H₂ + O₂ 75.3 % yield

H₂ + Cl₂ → 2HCl 69.8% yield

We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!

500.0gH_2O*\frac{1moleH_2O}{18.01 gH_2O}*\frac{2 moles H_2}{2 moles H_2O}*\frac{2.015g H_2}{1 mole H_2}*\frac{75.3 actual g}{100 theoretical g}=42.12 g H_2

42.12H_2*\frac{1 mole H_2}{2.015gH_2}*\frac{2 moles HCl}{1 mole H_2}*\frac{36.46g}{1 mole HCl}*\frac{69.8 actualg}{100 theoreticalg} =1064gHCl

Have a nice day!

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