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2 years ago

unlike magnetic strips, which store only account information, a ________ stores funds on an integrated circuit chip.

1 answer:

4 0

unlike magnetic strips, which store only account information, an E-pusre or e-wallet stores funds on an integrated circuit chip.

A digital wallet is an app used to carry out financial transactions on mobile devices (or electronic wallet).our passwords and payment details are securely stored there. Instead of carrying our cards with us while shopping, these programs let us pay using our mobile. we input and save the details of our credit card, debit card, or bank account so that we may use our smartphone to make purchases later on.

The purpose of digital wallets is to increase access to financial goods and services by utilizing the capabilities of mobile devices. Due to its ability to safely and conveniently store all of a user's payment information, digital wallets effectively do away with the need to carry a traditional wallet.

To know more about e-wallet visit : brainly.com/question/15994600

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Chapter 38, Problem 001

Norma-Jean [14]

Answer:

a) $\lambda=2.95x10^{-6}m$

b) infrared region

Explanation:

Photon energy is the "energy carried by a single photon. This amount of energy is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. If we have higher the photon's frequency then we have higher its energy. Equivalently, with longer the photon's wavelength, we have lower energy".

Part a

Is provide that the smallest amount of energy that is needed to dissociate a molecule of a material on this case 0.42eV. We know that the energy of the photon is equal to:

$E=hf$

Where h is the Planck's Constant. By the other hand the know that $c=f\lambda$ and if we solve for f we have:

$f=\frac{c}{\lambda}$

If we replace the last equation into the E formula we got:

$E=h\frac{c}{\lambda}$

And if we solve for $\lambda$ we got:

$\lambda =\frac{hc}{E}$

Using the value of the constant $h=4.136x10^{-15} eVs$ we have this:

$\lambda=\frac{4.136x10^{15}eVs (3x10^8 \frac{m}{s})}{0.42eV}=2.95x10^{-6}m$

$\lambda=2.95x10^{-6}m$

Part b

If we see the figure attached, with the red arrow, the value for the wavelenght obtained from part a) is on the infrared region, since is in the order of $10^{-6}m$