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d1i1m1o1n [39]
3 years ago
10

Give the value of the quantum number ℓ, if one exists, for a hydrogen atom whose orbital angular momentum has a magnitude of 6√(

h/2π).
Chemistry
1 answer:
MakcuM [25]3 years ago
5 0

Answer:

For this angular momentum, no quantum number exist

Explanation:

From the question we are told that

   The magnitude of the angular momentum is  L  = 6\sqrt{\frac{h}{2 \pi} }

The generally formula for Orbital angular momentum is mathematically represented as

           L  = \sqrt{(l * (l + 1)) } *  \frac{h}{2 \pi}

Where l is the quantum number

now  

We can look at the given angular momentum in this form as

      L  = 6\sqrt{\frac{h}{2 \pi} }    =  \sqrt{36}  * \sqrt{\frac{h}{2 \pi}} }

comparing this equation to the generally equation for Orbital angular momentum

     We see that there is no quantum number that would satisfy this equation

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List the first 20 elements chemistry
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An exciton is: A charged particle made of electrons A charged particle made of holes A neutral particle made of an electron and
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A gas mixture with 4 mol of Ar, x moles of Ne, and y moles
maks197457 [2]

Answer:

a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

b) x=4

c) \Delta G_{max}=-2721.9 J/mol

Explanation:

Gas mixture:

n_{Ar}= 4 mol

n_{Ne}= x mol

n_{Xe}= y mol

n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}

n_{Ne} + n_{Xe}=2*n_{Ar}

x + y=8 mol

y=8 mol- x

Mol fractions:

x_{Ar}=\frac{4 mol}{12 mol}=1/3

x_{Ne}=\frac{x mol}{12 mol}=x/12

x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12

Expression of \Delta G_{mixing}

\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

0=[ln (x/12)-ln ((8-x)/12)

ln (x/12)=ln ((8-x)/12)

x=(8-x)

x=4

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

4 0
3 years ago
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