B) Use the Rydberg equation to predict the first ionization...

b) Use the Rydberg equation to predict the first ionization energy of Li, of Na and of K. Remember to include the Z² term and use the appropriate value for Z. Show your work in the boxes below and enter your predicted ionization energy (to three significant figures) into the table below. Make sure to convert your energy into units of kJ mol?¹. ?E = IE = -RH · Z² (1/nf² - 1/ni²) (6.022x10²³/mol) Li z=3 -2.178 x 10?¹? J (3)² (1/?² - 1/(2)²) (6.022x10²³/mol) = 2.951 x 10? J/mol = 2.951 x 10³ kJ/mol Na z=11 ni=3 K z=19 ni=4

Transcript

00:02 Hello, in this question we need to use the rate -bur equation to find out the ionization energy.

00:13 So we know that that is rate -bar equation is equal to 1 by lambda is equal to rz square.

00:21 In bracket we have 1 by n1 square minus 1 by n2 square.

00:27 So therefore here this lambda, it is the wavelength.

00:32 Of the photon, r is the radberg constant which has the value 10973 -7 -1097 inverse, and z is the atomic number of an atom, n1 is the plank, this principal quantum number of lower energy level, and the n2 has the principal quantum number of the higher energy level.

01:04 That we ionized from the ground state.

01:07 So therefore n1 is equal to 1 and n2 will be since we are calculating the ionization energy.

01:14 So electron goes to the infinity with respect to the atom.

01:18 That is it leaves the atom here.

01:20 So therefore n2 will be infinity.

01:24 Now we have the lithium atom here.

01:27 So here the atomic number of lithium atom it is 3.

01:31 Now on putting the value we will have 1 by lambda is equal to r it is given that is 109737 multiply by z square the whole square of three in bracket we will have one divided by n1 is one square minus one divided by n2 is infinity square it is there so on solving this this bracket one will become zero because on division in the dividend part we have the in the denominator we have the infinity.

02:09 So, it will become now 109737 multiplied by 9 and hence we will have the 1 by lambda is equal to 987633 centimeter inverse.

02:25 Now we know that, that is e is equal to hc divided by lambda.

02:32 So on putting the value here, h is 6 .626 multiply, it is a plank constant.

02:38 Multiply by 10 to the power minus 34.

02:42 Multiply by c is the speed of light, that is 3, multiply by 10 raised to the power 8 divided by lambda is equal to 1 by 1 by lambda it is given.

02:55 So therefore lambda will be here 1 by 98763.

03:06 Or simply we can write it as 6 .6 to 6 multiply by 10 raised to the power minus 34, multiply by 3, multiply by 10 to the power 8.

03:15 This, the division part, it will come to the numerator in the form of multiplication.

03:23 9876 double 3.

03:25 So on solving this, we will have the value, the value of e is 1 .9632, multiply by 10 raised to the power minus 20 kilojoules...

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