Question

QUESTION 19 A solute is defined as the substance that is dissolved. the substance in which another substance dissolves. o the homogenous mixture of two or more species O the heterogeneous mixture of two or more species QUESTION 20 According to the Balmer Rydberg equation, electromagnetic radiation with wavelength 2. = 486,1 nm will be absorbed when an electron undergoes which of the following transition On-2-223 On-2--2-4 On - 3 -2 - 2 On-4-n2 = 2 QUESTION 21 Photochemists use electromagnetic radiation to initiate chemical reactions, often by providing the energy required to break bonds within a molecule. Lowering which of the following will result in electromagnetic radiation having more energy per photon? O amplitude O frequency intensity wavelength

Answer 1

Question 19

A solute is defined as substance that is dissolved.

Question 20

When light is absorbed by an atom, electron jumps from lower energy level to higher energy level. Therefore, n1= 3 to n2=2 and n1=4 to n2 =2 transitions are not possible.

Let's calculate wavelength associated with given electronic transitions.

we have equation $1585683637859_blob.png$ E = 2.18 $1585683638553_blob.png$ 10 -18 J [ ( 1 / n i2 ) - ( 1 / n f2 ) ]

Consider transition n = 2 $1585683638523_blob.png$ n = 3.

Here n i = 2 and n f = 3

$1585683638487_blob.png$$1585683638518_blob.png$E = 2.18 $1585683638491_blob.png$ 10 -18 J [ ( 1 / 2 2 ) - ( 1 /3 2 ) ]

$1585683638515_blob.png$E = 2.18 $1585683638525_blob.png$ 10 -18 J ( 0.139 )

$1585683638472_blob.png$E = 3.028  $1585683638825_blob.png$ 10 -19 J

We have relation, wavelength of emitted photon ( $1585683638808_blob.png$) = h c/ $1585683638820_blob.png$ E

Therefore, $1585683638520_blob.png$ = ( 6.626 $1585683638858_blob.png$ 10 -34 J s) ( 3.00 $1585683638849_blob.png$ 10 8 m/s ) / 3.028  $1585683638825_blob.png$ 10 -19J

$1585683638493_blob.png$ = 6.565 $1585683639255_blob.png$ 10 -7 m

We have relation, 1 m = 10 9 nm  

Therefore,  wavelength of photon =6.565 $1585683639255_blob.png$ 10 -7 m $1585683638811_blob.png$ ( 10 9 nm / 1 m )

wavelength of photon =656.5 nm

Consider transition n = 2 $1585683638523_blob.png$ n = 3.

Here n i = 2 and n f = 4

$1585683638487_blob.png$$1585683638518_blob.png$E = 2.18 $1585683638491_blob.png$ 10 -18 J [ ( 1 / 2 2 ) - ( 1 /42 ) ]

$1585683638515_blob.png$E = 2.18 $1585683638525_blob.png$ 10 -18 J ( 0.1875 )

$1585683638472_blob.png$E = 4.0875  $1585683638825_blob.png$ 10 -19 J

We have relation, wavelength of emitted photon ( $1585683638808_blob.png$) = h c/ $1585683638820_blob.png$ E

Therefore, $1585683638520_blob.png$ = ( 6.626 $1585683638858_blob.png$ 10 -34 J s) ( 3.00 $1585683638849_blob.png$ 10 8 m/s ) / 4.0875  $1585683638825_blob.png$ 10 -19J

$1585683638493_blob.png$ =4.863 $1585683639255_blob.png$ 10 -7 m

We have relation, 1 m = 10 9 nm  

Therefore,  wavelength of photon =4.863 $1585683639255_blob.png$ 10 -7 m $1585683638811_blob.png$ ( 10 9 nm / 1 m )

wavelength of photon =486.3 nm

ANSWER : n1= 2 to n2 = 4

Question 21

We have relation, Energy of photon ( E ) = h$\nu$ = hC / $\lambda$

where h is Plank constant, $\nu$ is frequency , C is velocity of light and $\lambda$ is wavelength.

From above relation, it is clear that   E $\propto$ $\nu$ and E $\propto$ ( 1/ $\lambda$ )

Hence, lowering wavelength of photon will increase energy of photon.

ANSWER : Wavelength