1) How many photons are produced in a laser pulse of 0.225 J at 6100 nm? ______

Question

1) How many photons are produced in a laser pulse of 0.225 J at 6100 nm?

______ photons

2) A line in the Brackett series of hydrogen has a wavelemgth of 26262 nm. From what state did the electron originated?

- ni= ___________

- In what region of the electromagnetic spectrum is this line observed? (visible,radio, x-rays, infrared, ultraviolet, microwaves, gamma rays)

3) A certain rifle bullet has a mass of 7.89g. Calcuate the de Broglie wavelength of the bullet traveling at 1361 miles per hour.

- ______m

4) What quantum numbers specify these subshells?

7s 5p 4d

n= _____ n=_____ n=______

l= _____ l=_____ l=_______

5) Complete this electron configuration for S.

- [Ne] ______

6) Complete this electron configuration for Sb.

- [Kr] _____________

Explanation / Answer

1)

n = ?

Given data:

Energy of n number of photons = 0.225 J

Wavelength () = 6100 nm = 6100 x 10-9 m = 6.1 x 10-6 m

And

c = 3.00 x 108 m/s and Planck’s constant = h = 6.626 x 10-34 J.s

Formula:

E = nhc/

n = E/hc

n = (0.225 x 6.1 x10-6) / (6.626 x 10-34x 3.00 x 108)

n = (1.3725 x 10-6)/(1.9878 x 10-25)

n = 6.9 x 1018

Number of photons = 6.9 x 1018

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2) For Brackett series of electromagnetic spectrum,

We have, n1 = 4 i.e. ni for Brackett series = 4

n2= ?

1/ = R[(1/42) – (1/n22)]

For Bracket series given data:

= 26262 nm = 26262 x 10-9 m = 26262 x 10-7 cm

R = Rydberg constant = 109678 cm-1

Put it in above equation and solve for n2

1/(26262 x 10-7) = 109678 [(1/16) - (1/n22)]

380.778 = 109678 [(0.0625) - (1/n22)]

[(0.0625) - (1/n22)] = 380.778/109678

[(0.0625) - (1/n22)] = 0.00347

1/n22 = 0.0625 – 0.00347

1/n22 = 0.05903

n22 = 16.9405

n2 = 4.12

Line will be observed in Infrared (IR) region.

==========================================

3)

De Broglie’s equation,

= h/mu

where,

= de Broglie wavelength in m

h = Planck constant = 6.626 x 10-34 J.s or Kg.m2/s

m = mass of particle = 7.89 g = 7.89 x 10-3 Kg

u = velocity of particle

u = 1361 miles/hr

u = 1361 x 1.60934 Km/hr ………..(1 mile = 1.60934 Km)

u = 2190 Km/hr

u = 2190 (1000 /3600) m/s ……….(1 Km/hr = 1000 m/3600 s)

u = 608.4 m/s

with these values in above equation,

= (6.626 x 10-34) /( 7.89 x 10-3 x 608.4)

= 1.38 x 10-34 m

de Broglie wavelength will be 1.38 x 10-34 m.

===========================================

4)

i) For 7s, n = 7 and l = 0

Explanation:

n = Principle quantum number and its 7 here

n = 7

We know that, for azimuthal quantum number ‘l’

Number of suborbitals = (2l + 1)

For s subshell number of suborbital = 1

Hence 2l+1 = 1

2l = 0

l =0

ii) For 5p, n = 5 and l = 1

Explanation:

n = Principle quantum number and its 5 here

n = 5

Now, Number of suborbitals = (2l + 1)

For p subshell number of suborbital = 3

Hence 2l+1 = 3

2l = 2

l =1

iii) For 4d, n = 4 and l = 2

Explanation:

n = Principle quantum number and its 4 here

We know that, for azimuthal quantum number ‘l’

Number of suborbitals = (2l + 1)

For d subshell, number of suborbital = 5

Hence 2l+1 = 5

2l = 4

l =2

===========================================

5)

Aufbau principle for electronic filling: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s ……..

S (atomic number 16)

[Ne] neon core account for 10 e and remaining 6 e filled by aufbau principle.

Electronic configuration S = [Ne] 3s2 3p4

===========================================

6) Sb (Atomic number 51)

[Kr] Krypton core account for 36 electrons and remaining 15 e filled by following aufbau principle.

Electronic configuration Sb: [Kr] 4d10 5s2 5p3

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