Question

6) Calculate the wavelength, in nanometers, of the light emitted by a hydrogen atom when its electron drops from the n the H atom are given by En-2.18 x 10-18 J (1/n2). (c = 3.00 x 108 m/s; h= 6.63 x 10-34 J. 7 to the n 4 principal energy level. Recall that the energy levels of s) A) 4.45 x 10-20 nm B)2.16 x 10-6 nm C) 9.18 x 10-20 nm D) 1.38 x 1014 nm E) 2.17 x 103 nm E-hv h n -2.16x10 9.170x10-20 -6425 C.63x1034 S

Answer 1

if electron transition from n=7 to n=4 principal energy level

then E_n= - 2,18_*10^-18 j( 1/n² )

  $\Delta$E=E₇ - E₄

$\Delta E$ = -2.18 * 10^-18 ( 1 / 4² - 1 / 7² )

=-2.18 * 10^-18 * 33 /784

=- 9,18 *10^-20

     

weknowAE = hv =

ΔΕ = hν =

hc X = ΔΕ

X= (6.626 * 10-34) (3.0 * 108)) 9.18 * 10-20

=2.1 *10^-6 m

the value of wavelength is =2.1 *10^-6 m