Question

if the lowest energy x rays have lambda = 3.055 × 10^-8 m, estimate the minimum difference in energy between two bohrs orbits such that an electronic transition would correspond to the emission of an x ray. assuming that the electrons in the other shells exert no influence , at what z (minm) would a transition from the second energy level to the first results in the emission of an x ray ?​

Answer 1

Answer:

$\large\boxed{\sf{z = 2 \; (Helium)}}$

Step-by-step explanation:

Given :

$\lambda = 3.055 \times {10}^{ - 8} \: m$

We know that

$\triangle E = \dfrac{hc}{ \lambda}$

where,

• h = planks constant
• c = speed of light

Putting the respective values, we get

$= > \triangle E = \dfrac{6.63 \times {10}^{ - 34} \times 3 \times {10}^{8} }{3.055 \times {10}^{ - 8} } \\ \\ = > \triangle E = 6.52 \times {10}^{ - 18} \: \: joule$

Also, we know that

$\triangle E_{H} = \dfrac{3}{4} (2.176 \times {10}^{ - 18} ) \: \: joule \\ \\ = > \triangle E_{H} = 1.63 \times {10}^{ - 18} \: \: \: joule$

Again, we know that

$\triangle E = \triangle E_{H}( {z}^{2} )$

$= > {z}^{2} = \dfrac{\triangle E}{\triangle E_{H}}$

Putting the respective values, we get

$= > {z}^{2} = \dfrac{6.52 \times {10}^{ - 18} }{1.63 \times {10}^{ - 18} } \\ \\ = > {z}^{2} = 4 \\ \\ = > z = 2$

Hence, value of z = 2 ( Helium )