The energy difference between the energy levels involved in the transition of the electron is directly proportional to the frequency of the emitted photon:

$\Delta E=h\nu(1)$

Where h is the Planck constant. The photon's frequency is inversely proportional to its wavelegth:

$\nu=\frac{c}{\lambda}(2)$

Here c is the speed of light. Replacing (2) in (1):

$\Delta E=\frac{hc}{\lambda}\\\Delta E=\frac{(6.63*10^{-34}J\cdot s)(3*10^8\frac{m}{s})}{121.6*10^{-9}m}\\\Delta E=1.64*10^{-18}J$

6 0

3 years ago

A 16 foot ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 3 feet per second, how fa

JulsSmile [24]

Answer:

11.625

Explanation:

L = length of the ladder = 16 ft

$v_{y}$ = rate at which top of ladder slides down = - 3 ft/s

$v_{x}$ = rate at which bottom of ladder slides

y = distance of the top of ladder from the ground

x = distance of bottom of ladder from wall = 4 ft

Using Pythagorean theorem

L² = x² + y²

16² = 4² + y²

y = 15.5 ft

Also using Pythagorean theorem

L² = x² + y²

Taking derivative both side relative to "t"

$0 = 2x\frac{\mathrm{d} x}{\mathrm{d} t} + 2y\frac{\mathrm{d} y}{\mathrm{d} t}$

$0 = x v_{x} + y v_{y}$

$0 = 4 v_{x} + (15.5) (- 3)$

$v_{x}$ = 11.625 ft/s

7 0

3 years ago