Chemistry, 25.03.2020 00:27, anna4060

The mathematical equation for studying the photoelectric effect is:

hν = W + 1/2 meμ^2 where ν is the frequency of light shining on the metal; W is the energy needed to remove an electron from the metal; and me and u are the mass and speed of the ejected electron, respectively. In an experiment, a student found that a maximum wavelength of 341 nm is needed to just dislodge electrons from a metal surface. Calculate the velocity (in m/s) of an ejected electron when the student employed light with a wavelength of 313 nm.

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Answers

The velocity of an ejected electron when the student employed light with a wavelength of 313 nm is $3.385\times 10^{5} m/s$ .

Explanation:

The mathematical equation for studying the photoelectric effect is:

$h\nu = W + \frac{1}{2} m_e\times \mu ^2$

We have :

Wavelength of light which can dislodge electrons from a metal surface = $\lambda =341 nm=341\times 10^{-9} m$

$1 nm=10^{-9} m$

Energy needed to remove an electron from the metal :

$W=\frac{hc}{\lambda }$ (Planck's equation)

$=\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{341\times 10^{-9} m}=5.829\times 10^{-19} J$

Frequency of light shinning on metal surface = $\nu =?$

Wavelength of light shinning on metal surface = $\lambda '=313 nm=313\times 10^{-9} m$

$nu=\frac{c}{\lambda '}=\frac{3\times 10^8 m/s}{313\times 10^{-9} m}=9.585 \times 10^{14}Hz$

Mass of electron = $m_e=9.109\times 10^{-31} kg$

Velocity of electron = u

$6.626\times 10^{-34} J.s\times 9.585\times 10^{14} Hz=5.829\times 10^{-19} J+\frac{1}{2}\times 9.109\times 10^{-31} kg\times (u)^2$

$u = 3.385\times 10^{5} m/s$

The velocity of an ejected electron when the student employed light with a wavelength of 313 nm is $3.385\times 10^{5} m/s$ .

answered: Guest

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

answered: Guest

Because electrons have negligible mass, to account for the mass of the isotope, there must be 16 neutrons, 16 neutrally charged, massive, fundamental particles present in the sulfur nucleus. and thus finally, if we have the 32s2− ion, there are 18 electrons, 16 protons, and 16 neutrons.