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LekaFEV [45]
3 years ago
12

A hydrogen atom requires a minimum energy of 2.18×10^-18 J atom to remove an electron from its ground state level. Determine whe

ther a blueviolet light with a wavelength of 434.0 nm can affect this process.
Chemistry
1 answer:
frosja888 [35]3 years ago
5 0
Just find the energy of the <span>blueviolet light with a wavelength of 434.0 nm using the formula:

E  = hc / lambda

E = energy
c= speed of light =  3 x 10^8 m/s
h = planck's constant =  6.6 x 10^{-34}  m^2 kg / s
lambda =  434 nm =  434 x 10^{-9} m

Putting these values (with appropriate units) in the above formula :

we get:  Energy, E = 4.5 x 10^{-19} J

E = 0.45 x 10^{-18} J

Now, the </span>minimum energy is 2.18×10^-{18} J but our energy is 0.45 x 10^{-18} J which is less.
<span>Means the electron will not be removed

</span>
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We know that:

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To calculate the time required, we use the equation:

I=\frac{q}{t}

where,

I = current passed = 40.0 A

q = total charge = 42551181 C

t = time required = ?

Putting values in above equation, we get:

40.0=\frac{42551181 C}{t}\\\\t=1063779sec

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, 1063779s\times \frac{1hr}{3600s}=295hr

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