B. 0.9 <span>
</span>D.Light intensity has no effect on whether electrons are emitted or not.
and
A. X=1.9eV,Y=0.2eV
I already took the gizmo so I know these are right. The first one I got wrong b/c there was no graph and the last one I didn't understand. Basically for the last one you calculate the work function for the metals and find their difference.
Hope this helps.
The total energy can be found by adding the different energies:
628 + 15,600 + 712
= 16.94 kJ
Answer:
9
Explanation:
The structure of fluorophore used in the experiments has been drawn in the attachment. And from the drawing counting we can say that there are 9 sp2-hybridized carbon atoms present. Fiuorophores are a fluorescent chemical compound that can re-emit light upon light excitation. Normally used to produce absorbance and emission spectra.
Answer:
ν = 7.04 × 10¹³ s⁻¹
λ = 426 nm
It falls in the visible range
Explanation:
The relation between the energy of the radiation and its frequency is given by Planck-Einstein equation:
E = h × ν
where,
E is the energy
h is the Planck constant (6.63 × 10⁻³⁴ J.s)
ν is the frequency
Then, we can find frequency,

Frequency and wavelength are related through the following equation:
c = λ × ν
where,
c is the speed of light (3.00 × 10⁸ m/s)
λ is the wavelength

A 426 nm wavelength falls in the visible range (≈380-740 nm)