Question

1. A 100-watt lightbulb radiates energy at a rate of 100 J/s (The watt, a unit of power, or energy over time, is defined as 1 J/s). If all of the light emitted has a wavelength of 505nm, how many photons are emitted per second? (Assume three significant figures in this calculation.)

2.Calculate the de Broglie wavelength of a 143-g baseball traveling at 80 mph .

3.An electron undergoes a transition from an initial (ni) to a final (nf ) energy state. The energies of the ni and nf energy states are −2.179×10⁻¹⁸ J and −8.720×10−20 J, respectively.

Calculate the wavelength (λ) of the light in nanometers (nm) corresponding to the energy change (ΔE) value of this transition. You can use the following values for your calculations:

Planck′s constant (h)speed of light (c)1 m===6.626×10−34 J⋅s2.998×108 m/s109 nm

Answer 1

Q. 1)

Power = 100 w (J/s)

Wavelength of a photon = 505 nm = 505 E -9 m

Calculation of energy of a photon

E = hc/l = sf

Here E is energy in J , h is planks constant = 6.626 E-34 ,

c = speed of light = 3.0 E8 m/s

l = is wavelength in m

E = 6.626 E-34 Js x 3.0 E8 m/s )/ 505 E-9 m

= 3.94 E-19 J

Calculation of number of photon

Emission of energy per s = 100 J

Number of photon in 100 J = 100 J x 1 photon / 3.94 E-19 J

= 2.54 E20 photon

Since the energy emission is for one second so photon emitted per second

= 2.54 E20 photon

Q. 2 )

            l = h/P

Here l = is wavelength in m , h is planks constant

P : momentum = mv

Here m is mass in kg and v is velocity in m/s

Calculation of mass in kg

= 143 g x 1 kg / 1000 g

= 0.143 kg

V = 80 m/hr

Conversion of velocity to m/s

= (80 m/ hr )x (1hr/3600 s)

= 0.022 m/s

l = 6.626 E-34 J per s / (0.143 kg x 0.0222 m/s)

= 2.085 E-31 m

3.

l = hc / Delta E

Delt a E = E(final) – E(initial)

l = 6.626 E-34 Js x 2.998 E8 m per s )/ [ ( -8.720 E-20 J ) – (- 2.179 E-18 ) ]

= 9.50 E-8 m

l = 9.50 E-8 m