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maria [59]
3 years ago
5

A ground state hydrogen atom absorbs a photon of light having a wavelength of 92.05 nm. It then gives off a photon having a wave

length of 1736 nm. What is the final state of the hydrogen atom?
Chemistry
1 answer:
vodka [1.7K]3 years ago
4 0
Absorbed photon energy
Ea = hc/λ.. (Planck's equation)
Ea = hc / 92.05^-9m 

<span>Energy emitted
Ee = hc/ 1736^-9m </span>

Energy retained ..
∆E = Ea - Ee = hc(1/92.05<span>^-9 - 1/1736^-9) </span>
<span>∆E = (6.625^-34)(3.0^8) (1.028^7)
∆E = 2.04^-18 J </span>

<span>Converting J to eV (1.60^-19 J/eV)
 ∆E = 2.04^-18 / 1.60^-19
∆E = 12.70 eV </span>

<span>Ground state (n=1) energy for Hydrogen = - 13.60eV </span>

<span>New energy state = (-13.60 + 12.70)eV = -0.85 eV </span>

<span>Energy states for Hydrogen
En = - (13.60 / n²) </span>

n² = -13.60 / -0.85 = 16
n = 4
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The units used to measure specific heat capacity is Joules per kilogram per Kelvin.

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Nuclear energy comes from an atom's _____. <br><br> A. shells B. charge C. nucleus D. electrons
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5 0
3 years ago
How many hydrogen atoms are present in .46 moles of NH3
mixas84 [53]

Answer:

2.78 x 10²³

Explanation:

1 mole contains 6.02 x 10²³ hydrogen atoms => 0.46 mole contains 0.46(6.02 x 10²³) hydrogen atoms or 2.78 x 10²³ atoms.

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3 0
3 years ago
Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of
lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

U_{2}-U_{1}=Q-W

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

dw=Pdv

The pressure is constant, so:  w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg} (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

U_{2}-U_{1}=500-(3*75)=275kJ

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

P_{1}V_{1}=nRT_{1}

P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}  

Subtracting the first from the second:

1.5P_{1}V_{1}=nR(T_{2}-T_{1})

Isolating T_{2}:

T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}

Assuming that it is water steam, n=0.1666 kmol

V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}

T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76 ºC

7 0
3 years ago
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