Question

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Answer 1

Answer:

your answer in the picture ...

hope you get it...

Answer 2

ohk,, don't scare,,I am completing whole answer

1) As we know, acccording to de-brogile equation

wavelength=h/mv

simply putting the value of wavelength and h,we get

momentum=2.2×10^-11 ×6.6×10^-34

=1.45×10^-44

2)optiond) ✔️✔️

3)Optionb✔️✔️

3)Optionb✔️✔️4) optionc)✔️✔️

tionc)✔️✔️reason:-As we know

E=-13.6×z^2/n^2

due to negative sign,we can say less is the n , less is the energy

so,

energy will be minimum at orbital=1

to move from first to other orbital ,it absorbs energy

to move from other orbital to first ,it loss energy

5) increases

5) increasesReason:-similar to above

6)The emmision of line spectrum,from other orbits(greater than 2) to second orbit is balmer series

for the third line

The lines should come from fifth orbit to second one

hence optiona)

7) Hiesenberg uncertainty principal

8)As we know

1/wavelength=R(1/(n1)^2 -1/(n2)^2)

here

R is constant which is eqaul to 10976cm

1/wavelength=10976×(1-1/4)cm

=2744×3cm

wavelength=1/8232×10^-2m

now as we know

frequency=c/wav.

frequency=3×10^10×1083/

=30.57×10^10(approx)

9) using the de-brogile wavelength as like 1) we get

wavelength=h/mv

here wavelength=1.54×10^-10m

v=3×10^8m/s

h=6.626×10-34

m=(3×10^8×1.54×10^-10)/6.626×10^-34

m=(4.62×10^32)/6.6

m=.8628×10^-34

(numerical questions are covered now)