Kwant światła fioletowego o dł fali 412,5nm ma energię...
a. ½ eV
b.2 eV
c.⅓ eV
d.16 eV
+obliczenia
E= h * f
f= c/λ
1eV= 1,6 *10 ⁻¹⁹J
h= 6.63 * 10 ⁻³⁴ Js
c= 3* 10⁸m/s
412,5nm= 412,5* 10⁻⁹ m
E= 6.63 * 10 ⁻³⁴ Js* (3* 10⁸m/s / 412,5* 10⁻⁹ m)
E= 6.63 * 10 ⁻³⁴ Js * 7.3* 10¹⁴ 1/s
E≈ 4,8* 10⁻¹⁹J
E= 3eV
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
E= h * f
f= c/λ
1eV= 1,6 *10 ⁻¹⁹J
h= 6.63 * 10 ⁻³⁴ Js
c= 3* 10⁸m/s
412,5nm= 412,5* 10⁻⁹ m
E= 6.63 * 10 ⁻³⁴ Js* (3* 10⁸m/s / 412,5* 10⁻⁹ m)
E= 6.63 * 10 ⁻³⁴ Js * 7.3* 10¹⁴ 1/s
E≈ 4,8* 10⁻¹⁹J
E= 3eV