#12 Sodium metal has a work function of 3.78 #215; 10^-19 J. Does light of 585 nm have suffic

Step 1: Calculate the energy of the light with a wavelength of 585 nm using the formula E = hc/λ

#12 Sodium metal has a work function of 3.78 #215; 10^-19...

#12 Sodium metal has a work function of 3.78 × 10^-19 J. Does light of 585 nm have sufficient energy to trigger the photoelectric effect of sodium? E = Energy of photon h = plank's constant (6.626 × 10^-34 Js) ? = wavelength of photon c = velocity of light E = 3.78 × 10^-19 J ? = h × c / E = (6.626 × 10^-34 × 3 × 10^8) / 3.78 × 10^-19 ? = 5.2587 × 10^-7 m or ? = 526 nm #20 How many photons of light are emitted when an electron falls from a. the fourth energy level to the first in a hydrogen atom? b. the fourth energy level to the third and then to the first energy level? c. Compare the total energy released in parts (a) and (b).

Transcript

00:03 In the 12th question, it is asked, does light of 585 nanometer have sufficient energy to trigger the 4 to electric representative sodium? so here, e is equal to hc by lambda, therefore lambda is equal to xc by e.

00:19 So, value of h, it is a plant's constant is, it's 0 .66 into 10 power minus 34, june 2nd multiplied by the value of speed of light is 3 into 10 power 8 mid -per -second, divided by the energy is 3 .78 into 10 power minus 19 joules, which is equal to 5 .2587 into 10 power minus 7 meter.

00:46 Convering this into nanometer, which is equal to 525 .87 nanometer.

00:58 So 585 nanometer have not, does not have the sufficient energy because energy is inversely proportional to 1 by lambda.

01:26 Now coming to the 20th question a part, it is asked the fourth energy level to the first in the hydrogen atom.

01:42 So 1 by lambda is equal to r into 1 by n square final minus 1 by n square emission r is a ritzburg's constant which values 1 .097 into 107 meter so which is equal to 1 .097 into 10 7 per meter into 1 by 1 square minus 1 by 5 square which is equal to 1 .097 into 0 .097 into 0 .90 this is 1 by lambda so lambda is equal to 1 by 1 .0 .7 into 0 .9 6 this is 1 by lambda so lambda is equal to 1 by 1.

02:28 097 into 10 power 7 into 0 .96, which is equal to 9 .50 into 10 power minus 8 meter.

02:44 And the energy is equal to hc by lambda, which is equal to 6 .66 into 10 power minus 34, multiplied by 3 into 10 power 8 by lambda is equal to 9 .50 into 10 minus 8 meter, which is equal to 2 .50.

03:06 0 .09 into 10 power minus 18 joules.

03:14 Coming to the b part, same formula...

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