Help me with this please. :))

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

Which graph shows the correct relationship between kinetic and speed

Elenna [48]

I can’t answer without any graph options

6 0

2 years ago

Read 2 more answers

Consider two waves X and Y traveling in the same medium. The two carry the same amount of energy per unit time, but X has one-se

RideAnS [48]

Answer:

7 / 1

Explanation:

The ratio of their amplitude = one-seventh and the ratio of their amplitude = the ratio of their wavelength

Ax / Ay = λx / λy = 1 / 7

λy / λx = 7 / 1

7 0

3 years ago

The data table shows the cause of unintentional death for various age groups; what can you infer about the 15-24 age group?

Blizzard [7]

Answer:

D) They most likely died from not wearing a seatbelt.

Explanation:

Their death was caused by a "motor vehicle" (that's what MV stands for in this case). The most logical answer would be D.

3 0

3 years ago

Read 2 more answers

The angular momentum of a flywheel having a rotational inertia of 0.140 kg ·m2 about its central axis decreases from 3.00 to 0.8

Rasek [7]

Answer

given,

I = 0.140 kg ·m²

decrease from 3.00 to 0.800 kg ·m²/s in 1.50 s.

a) $\tau = \dfrac{\Delta L}{\Delta t}$

$\tau = \dfrac{0.8-3}{1.5}$

τ = -1.467 N m

b) angle at which fly wheel will turn

$\theta= \omega t +\dfrac{1}{2}\alpha t^2$

$\theta= \dfrac{L}{I} t +\dfrac{1}{2}\dfrac{\tau}{I}t^2$

$\theta= \dfrac{3}{0.14}\times 1.5+\dfrac{1}{2}\dfrac{-1.467}{0.14}\times 1.5^2$

θ = 20.35 rad

c) work done on the wheel

W = τ x θ

W = -1.467 x 20.35 rad

W = -29.86 J

d) average power of wheel

$P_{av} =-\dfrac{W}{t}$

$P_{av} =-\dfrac{(-29.86)}{1.5}$

$P_{av} =19.91\ W$

7 0

3 years ago

Read 2 more answers