What is the energy carried by a 600 nm photon in eV? h = 6.626 ×10 34 Js, 1 eV =

Question

What is the energy carried by a 600 nm photon in eV? h = 6.626 ×1034 Js, 1 eV = 1.602 x1019 J.

2.07 eV.

1.01 eV.

5.12 eV.

3.12 eV.

How does the atomic radius vary across one row of the periodic table in the increasing Z direction? Why?

a)it decreases because the quantnum number n of the outer electron decreases

b)it decreases, because the increasing effective nuclear charge pulls the electrons in closer
c)it increases, because the quantum number n of the outer electron increases.
d)it  increases, because the number of electrons increases.

2.07 eV.

1.01 eV.

5.12 eV.

3.12 eV.

Explanation / Answer

E = hf, where h = Plank's constant 6.626 x 10^-34Js
E = hc/
E is the energy of the particle in Joules
f is frequency in Hertz, is wavelength in meters
c is speed of light in m/s

now wavlength=600nm = 600 x10-9m

so E =( 6.626 x10-34Js) (3x108 m/s) / (600 x10-9m)

E= 3.313 X 10-19 joules

Now 1 eV = 1.602x 10-19 J

so E= 3.313 X 10-19 /1.602 x 10-19

E=2.07 ev

so option (a) is correct.

(2)As we move in a period, the quantam number n remains same but the effective nuclear charge increases continously which pulls the electrons closer to nucleus and hence the atomic radius decreases. so option b is correct

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