Question

Calculate the frequency of the light emitted when an electron in a hydrogen atom makes each of the following transitions. Express in inverse seconds.

a) n=4?n=3

b) n=5?n=1

c) n=5?n=4

d) n=6?n=5

Answer 1

usefull formulas

E = (-2.18 * 10^-18 ) [1/(n₂)² - 1/(n₁²] Jouls

n2 = higher value , n1 =lower value

E = h v

h = 6.625 x 10^-34 J S

a ) solution :

E = (-2.18 * 10^-18 ) [1/(n₂)² - 1/(n₁²]

        = (-2.18 * 10^-18 ) [1/(4)² - 1/( 3²)]

         = 1.06 x 10^-19

E = h v

v = E / h

   = 1.06 x 10^-19 / 6.625 x 10^-34

   = 1.60 x 10^14 sec^-1

frequency = 1.60 x 10^14 sec^-1

b ) solution :

E = (-2.18 * 10^-18 ) [1/(n₂)² - 1/(n₁²]

        = (-2.18 * 10^-18 ) [1/(5)² - 1/( 1²)]

         = 2.09 x 10^-18

E = h v

v = E / h

   = 2.09 x 10^-18 / 6.625 x 10^-34

   = 3.16 x 10^15 sec^-1

frequency = 3.16x 10^15 sec^-1

c ) solution :

E = (-2.18 * 10^-18 ) [1/(n₂)² - 1/(n₁²]

        = (-2.18 * 10^-18 ) [1/(5)² - 1/( 4²)]

         = 4.90 x 10^-20

E = h v

v = E / h

   = 4.90 x 10^-20 / 6.625 x 10^-34

   = 7.40 x 10^13 sec^-1

frequency = 7.40x 10^13 sec^-1

d ) solution :

E = (-2.18 * 10^-18 ) [1/(6)² - 1/(5²]

        = (-2.18 * 10^-18 ) [1/(5)² - 1/( 4²)]

         = 2.66 x 10^-20

E = h v

v = E / h

   = 2.66 x 10^-20 / 6.625 x 10^-34

   = 4.02 x 10^13 sec^-1

frequency = 4.02x 10^13 sec^-1